I have been reading Anand Pillay's An Introduction to Stability Theory. In Lemma 1.15, he constructs from a type $p(\bar x)$ over a model $M$ a type over $A \supset M$ that is an heir of $p$.
The argument goes like this: If $\Sigma (\bar x) := \mathrm{Th}_{L(A)}(\mathbf M) \cup p(\bar x) \cup \{\neg\phi(\bar x, \bar a) \mid \phi \in L(M), \bar a \in A, \text{$\phi(\bar x, \bar y)$ is not represented in $p(\bar x)$}\}$, where $\mathbf M$ is a big (monster) model, is "consistent," then a completion of $\Sigma$ is an heir of $p$. (I understand that in this book, and perhaps in model theory in general, "consistent" means "consistent with the elementary diagram of $\mathbf M$.") Of course, we use compactness to see the consistency of $\Sigma$. An arbitrary finite subset of $\Sigma(\bar x)$ is without loss of generality a formula $\Gamma$ of the form $\chi(\bar m, \bar a, \bar b) \land \alpha(\bar x, \bar m) \land \neg\phi_1(\bar x, \bar a, \bar m) \land \dots \land \neg\phi_k(\bar x,\bar a, \bar m)$, where $\bar m \in M$, $\bar a, \bar b \in A \setminus M$, $\chi(\bar m, \bar a, \bar b) \in \mathrm{Th}_{L(A)}(\mathbf M)$, and $\phi_j(\bar x, \bar y, \bar m)$ is not represented in $p(\bar x)$. Since $M$ is an elementary submodel of $\mathbf M$, there are $\bar a', \bar b' \in M$ s.t. $\models \chi(\bar m, \bar a', \bar b')$. It can be shown that if you replace $\bar a \bar b$ with $\bar a'\bar b'$ in $\Gamma$, the resulting formula is in $p(\bar x)$, hence $\Gamma$ is consistent.
What I do not understand is why we can replace $\bar a \bar b$ with $\bar a'\bar b'$ to conclude the consistency of $\Gamma$. It seems to me that to use that argument the types of $\bar a\bar b$ and $\bar a' \bar b'$ must be the same, especially given the point in parentheses above, but nothing seems to guarantee this. I'm not used to arguments that involve monster models; what am I missing?
As you've noted, in Chapter 0 of Pillay's book (on p. 7), he says that for a set $\Sigma$ of formulas with parameters from a subset $A$ of the monster model $\overline{M}$, when we say "$\Sigma$ is consistent", we really mean "$\Sigma\cup \text{Th}(\overline{M},a)_{a\in \overline{M}}$ is consistent". Then he notes immediately that this is equivalent to saying "$\Sigma\cup \text{Th}(\overline{M},a)_{a\in A}$ is consistent". In other words, $\Sigma$ is consistent with the elementary diagram of $A$. I prefer the latter formulation.
Why should this be the meaning of "$\Sigma$ is consistent"? Well, if $\Sigma\cup \text{Th}(\overline{M},a)_{a\in A}$ is satisfied in a model $M$ of $T$, then the interpretation of $A$ in $M$ has the same elementary diagram as the "real" $A$ in $\overline{M}$, so by saturation $M$ embeds in $\overline{M}$ over $A$, and we've actually realized $\Sigma$ over the parameters $A$. On the other hand, if we were only show consistency of $\Sigma$ on its own, the constant symbols from $A$ might be interpreted as elements which look very different from the "real" $A$.
Note that in your question, this convention of including the elementary diagram of $A$ is actually made explicit in $\Sigma(\overline{x})$ as you wrote it.
Now when you actually try to use compactness to prove that $\Sigma\cup \text{Th}(\overline{M},a)_{a\in A}$ is consistent, you have two choices. You can show that any finite subset $\Sigma'$ of $\Sigma$ is consistent with $\text{Th}(\overline{M},a)_{a\in A}$ by finding a realization of $\Sigma'$ over $A$ in the monster model. OR you can show that any finite subset of $\Sigma(\overline{x})\cup \text{Th}(\overline{M},a)_{a\in A}$ is consistent, with nothing in the background. Here, the parameters from $A$ are just constant symbols, and we have no obligation to satisfy the whole of the elementary diagram of $A$, only a finite piece of it!
The latter kind of argument is what Pillay is referring to when he writes "Of course, this is also equivalent to saying that every finite subset $\Sigma'$ of $\Sigma\cup \text{Th}(\overline{M},a)_{a\in A}$ is consistent (with $\emptyset$)... In a few cases we use compactness in this latter sense."