I'm having a bit of trouble with Lemma 6.3.2. of Tent and Ziegler's model theory textbook. Here is the lemma and their proof, transcribed verbatum.
Lemma. Let $T$ be $\omega$-stable, $M\prec N$ models of $T$, $\varphi(v)$ be strongly minimal and $b_i\in\varphi(N)$. If the $b_i$ are independent over $\varphi(M)$, they are independent over $M$.
Proof: Assume that $b_1,\ldots,b_n$ are algebraically independent over $\varphi(M)$ but dependent over $\bar{a}\in M$. Put $\bar{b}=(b_1,\dots,b_n)$. An argument as in the proof of Theorem 5.5.2 shows that we may assume that $M$ is $\omega$-saturated. Let $p$ be the type of $\bar{b}$ over $M$. We choose a sequence $\bar{b}^0,\bar{b}^1,\dots$ in $\varphi(M)$ such that $\bar{b}^{2i}$ is an $n$-tuple of elements algebraically independent over $\bar{b}^0,\dots,\bar{b}^{2i-1}$ and $\bar{b}^{2i+1}$ realizes $p\upharpoonright_{\bar{a},\bar{b}^0,\dots,\bar{b}^{2i}}$. Let $q$ be the type of $\bar{a}$ over the set $B$ of elements of $(\bar{b}^i)$. Since the sequence $(\bar{b}^i)$ is indiscernible, every permutation $\pi$ of $\omega$ defines a type $\pi(q)$ over $B$. If $\{i:\pi(2i)\text{ even}\}=\{i:\pi'(2i)\text{ even}\}$, we have $\pi(q)\neq\pi'(q)$. So there are uncountably many types over $B$ and $T$ is not $\omega$-stable.
Unfortunately, a few points of this proof are unclear to me. First, I think I have a sketch of a proof that we may assume $M$ is $\aleph_0$-saturated; we essentially enrich our language by including a single unary predicate symbol, call it $U$, such that $U(N)=M$. There are sentences expressing that $U$ is a proper elementary substructure, and we may further form a set of formulas over free variables $v_1,\dots,v_n$ expressing that $(v_1,\dots,v_n)$ is algebraically independent over $\varphi(U)$, by taking formulas of the form $$\forall\bar{w}(\bigwedge_j(\varphi(w_j)\wedge{U(w_j)})\rightarrow((\exists^{\leqslant k}\bar{u}\theta(\bar{u},\bar{w}))\rightarrow\neg\theta(\bar{v},\bar{w})))$$ for all $k\in\omega$ and $\theta\in\operatorname{Form}(\mathcal{L})$. Then $(b_1,\ldots,b_n)$ satisfies these formulas in $N$, and so if we take the union of an elementary chain $(N_i)_{i\in\omega}$ such that $N_0=N$ and all types over finite subsets of $U(N_i)$ are realized in $U(N_{i+1})$, we obtain a new structure $N'$ such that $M':=U(N')$ is $\aleph_0$-saturated and $(b_1,\dots,b_n)$ is not algebraic over $\varphi(M')$. Clearly $(b_1,\dots,b_n)$ will still be dependent over $\bar{a}\in M'$, witnessed by the same formula as in $N$, and so since $M'\prec N'$ we have that all the hypotheses of the theorem are satisfied. So, first question, is this sketch the right idea?
It's at this point that I start to get confused. Constructing the sequence $(\bar{b}^i)$ seems clear to me; the case where $i$ is odd is immediate, and for the even case we can do a similar trick as above, taking the partial type $$\Sigma(\bar{v})=\{\bigwedge_{i=0}^n\varphi(v_i)\}\cup\{(\exists^{\leqslant k}\bar{u}\theta(\bar{u}))\rightarrow\neg\theta(\bar{v}):\theta\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})\}_{k\in\omega}.$$ If this were not finitely satisfiable, there would be algebraic formulas $\theta_1,\ldots,\theta_m\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})$ such that $$M\models\forall\bar{v}(\bigwedge_{i=0}^n\varphi(v_i)\rightarrow\bigvee_{j=1}^m\theta(\bar{v})),$$ which would mean $\varphi(M)^n$ is finite, an obvious contradiction. Thus $\Sigma(\bar{v})$ is consistent and so realized in $M$. Is this right?
However, I have absolutely no idea why the sequence $(\bar{b}^i)$ is indiscernible. We wish to show that, for any $\mathcal{L}$-formula $\psi(\bar{v}_1,\dots,\bar{v}_m)$, and any $i_1<\dots i_m\in\omega$ and $j_1<\dots j_m\in\omega$, we have $$M\models\psi(\bar{b}^{i_1},\dots,\bar{b}^{i_m})\leftrightarrow \psi(\bar{b}^{j_1},\dots,\bar{b}^{j_m}).$$ Certainly it suffices to show the $\rightarrow$ implication, and we clearly need to divide into cases depending on whether $i_m$ and $j_m$ are odd or even. My attempt at this has been with induction on $m$; for instance, the case $m=0$ is vacuous. For the case $m=1$, we first consider the case where $i_1$ and $j_1$ are both odd. Then $$M\models \psi(\bar{b}^{i_1})\iff\psi(\bar{v})\in p\upharpoonright_{\emptyset}\iff M\models \psi(\bar{b}^{j_1}),$$ as needed. Next we might consider the case where $i_1$ and $j_1$ are both even, but even here I seem to get stuck. We know that $M\models \psi(\bar{b}^{i_1})$ forces $\psi(\bar{v})$ to be non-algebraic, but this alone is certainly not enough to conclude that $M\models\psi(\bar{b}^{j_1})$. I imagine that we want to use strong minimality of $\varphi$ here somehow, to show that if $M\models\neg\psi(\bar{b}^{j_1})$ then $\bar{b}^{j_1}$ would be algebraic, but how can we exploit this when $\psi$ has more than one free variable? Indeed, $\bigwedge_{i=1}^n\varphi(v_i)$ will not be a strongly minimal formula when $n>1$ (for instance, consider its intersection with the formula $v_1=c$ for any $c\in \varphi(M)$, which is infinite and coinfinite), and I don't see an obvious way to "extract" a strongly minimal formula in $n$ free variables from a strongly minimal formula in $1$ free variable. So I'm not sure how to proceed even in the case $m=1$, let alone cases where $m>1$. I get even further stuck when $i_m$ and $j_m$ have different parities, as I can't see a way to link the construction of odd entries in the sequence with even entries in the sequence.
Furthermore, even if the $(\bar{b}^i)$ were indiscernible, why would this mean that $\pi(q)$ is a type? The only thing I can imagine they might mean by $\pi(q)$ is $$\{\psi(\bar{v},\bar{b}^{\pi(i_0)},\dots,\bar{b}^{\pi(i_n)}):\psi(\bar{v},\bar{b}^{i_0},\dots,\bar{b}^{i_n})\in q\},$$ but for general sequences of indiscernibles I can't see why this construction would possibly yield types. For instance, consider $(\mathbb{Q},<)$ and any infinite sequence of distinct elements $B\subseteq\mathbb{Q}$. If $b_1<b_2\in B$, then any type over $B$ will contain the sentence $b_1<b_2$, so the "type" induced by any permutation $\pi:B\rightarrow B$ that swaps $b_1$ and $b_2$ will contain the sentence $b_2<b_1$ and hence not be consistent. I'm clearly missing something here; what do they actually mean by $\pi(q)$?
Q1: Yes, your sketch of why we may assume $M$ is $\aleph_0$-saturated is correct.
Q2: Yes, your construction of $\bar{b}^i$ when $i$ is even is correct. But there is a typo here in Tent and Ziegler: $\bar{b}^i$ should be chosen to be algebraically independent over $\bar{a}, \bar{b}^1,\dots,\bar{b}^{i-1}$, not just $\bar{b}^1,\dots,\bar{b}^{i-1}$ (so that $\text{tp}(\bar{a}/\bar{b}^i)$ contains a formula witnessing algebraic dependence of $\bar{b}^i$ over $\bar{a}$ when $i$ is odd, but no such formula when $i$ is even).
Q3: Why is the sequence $(\bar{b}^i)_{i\in \omega}$ indiscernible? Well, it comes down to the fact that in a strongly minimal set, for each $m$, there is a unique type satisfied by any algebraically independent $m$-tuple (this is Corollary 5.7.4 in Tent and Ziegler). Now for any $i_1<\dots<i_m$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, so its type does not depend on the choice of $i_1<\dots<i_m$.
Note that the distinction between the odd and even cases has disappeared, since we're just asking for indiscernibility (over $\varnothing$), not indiscernibility over $\bar{a}$. If we forget about $\bar{a}$, each new $\bar{b}^i$ is just a tuple which is algebraically independent over $\bar{b}^1,\dots,\bar{b}^{i-1}$. In the even case, we explicitly chose it that way, and in the odd case, that's part of the type $p\restriction \bar{b}^1,\dots,\bar{b}^{i-1}$, since $\bar{b}^1,\dots,\bar{b}^{i-1}\in \varphi(M)$ and $\bar{b}$ is algebraically independent over $\varphi(M)$.
Some superfluous information: This kind of indiscernible sequence, where each element of the sequence is chosen to be independent over the previous elements, is called a Morley sequence. The concept of Morley sequence, and especially its generalizations outside the strongly minimal context (where we need to use more general notions of independence than algebraic independence - most commonly non-forking independence) are hugely important in model theory.
Q4: Your understanding of what is meant by $\pi(q)$ is correct. So why is $\pi(q)$ a type? The crucial thing that Tent and Ziegler have neglected to tell you here is that $(\bar{b}^i)_{i\in \omega}$ is not just an indiscernible sequence, it's an indiscernible set. This means that for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, and any other tuple $j_1,\dots,j_m$ of pairwise distinct elements of $\omega$, $\text{tp}(\bar{b}^{i_1},\dots,\bar{b}^{i_m}) = \text{tp}(\bar{b}^{j_1},\dots,\bar{b}^{j_m})$. This is just the same as the definition of indiscernible sequence, but we no longer require $i_1,\dots,i_m$ and $j_1,\dots,j_m$ to be increasing tuples, i.e., we forget about the ordering of the sequence. An indiscernible set is also called a totally indiscernible sequence - see the definition at the beginning of Section 9.1 of Tent and Ziegler.
It follows that for any permutation $\pi$ of $\omega$, $\text{tp}((\bar{b}^i)_{i\in \omega})) = \text{tp}((\bar{b}^{\pi(i)})_{i\in \omega})$, and thus if $p$ is a consistent type over $(\bar{b}^i)_{i\in \omega}$, then $\pi(p)$ (which I would denote by $\pi_*p$) is also consistent (by the lemma on pushforward types, which we've already discussed here).
Ok, so why is $(\bar{b}^i)_{i\in \omega}$ an indiscernible set? The key point here is that algebraic independence is a symmetric notion: if $a_1,\dots,a_k$ is a tuple of algebraically independent elements, then so is $a_{\sigma(1)},\dots,a_{\sigma(k)}$ for any permutation $\sigma$ of $\{1,\dots,k\}$. So for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, and this does not depend on the choice of tuple $i_1,\dots,i_m$ (and in particular not on its ordering).
Some more superfluous information: