Going through Hodge's Shorter Model Theory, there is one thing I'm finding a bit confusing.
Two L-structures are elementarily equivalent when they satisfy precisely the same first-order sentences. Fine. But you have situations where you have models A and B for the same theory, some tuple $\bar{a}$ in A, another tuple $\bar{b}$ in B, and end up talking about $(A,\bar{a})$ being elementarily equivalent to $(B,\bar{b})$ (or perhaps only satisfying the same quantifier-free sentences, as in the hypothesis of Theorem 7.4.1 (b)).
Does it even make sense to say something like $(A,\bar{a}) \equiv (B, \bar{b})$? Surely the first is an $L(\bar{a})$-structure, whereas the second is an $L(\bar{b})$-structure... What gives?
If $L=\{\leq\}$, for example, can we even ask whether $(\mathbb{Q}, <, \frac{2}{3}) \equiv (\mathbb{Z}, <, 6)$? We can ask if the first structure satisfies the sentence $\exists x\ (x<\frac{2}{3})$ in $L(\frac{2}{3})$, but not if the second one does, since it's not $L(\frac{2}{3})$-structure, so how could they satisfy exactly the same sentences...? Similarly for quantifier-free sentences.
One thought would be to just assume that we're implicitly expanding both languages to some common $L(\bar{c})$, where $\bar{c}$ includes both $\bar{a}$ and $\bar{b}$, but then you're left with the problem of having to interpret the constants $\bar{b}$ in A and $\bar{a}$ in B, so I'm confused.
Your last paragraph is moving in the right direction, but is mixing up elements and constant symbols: when we write $$(A,\overline{a})\equiv(B,\overline{b})$$ we mean that the expansions of $A$ and $B$ by a tuple of new constant symbols $\overline{c}$ interpreted as $\overline{a}$ and $\overline{b}$ respectively are elementarily equivalent. Note that the tuple $\overline{c}$ has the same length as each of the tuples $\overline{a}$ and $\overline{b}$!
A key point here is that tuples are ordered (so there's no ambiguity about how to assign the new symbols): we are really writing the above as an abbreviation for $$(A,a_1,a_2,...,a_n)\equiv (B,b_1,b_2,...b_n),$$ and so we're letting our $i$th new constant symbol refer to $a_i$ in the expansion of $A$ and $b_i$ in the expansion of $B$. Note that this means that order matters: for instance, if $A=B=(\mathbb{R};+,\times,<)$, then $$(A, e,\pi)\not\equiv (A,\pi,e)$$ (consider the sentence "$c_1<c_2$," which is true in the former but not the latter).
What if for some reason you want to consider "expansions by unordered tuples"? Well, then you'd use unary relation symbols: given $\{a_1,...,a_k\}\subseteq A$ and $\{b_1,...,b_k\}\subseteq B$ (and conflating $A$ and $B$ with their underlying sets for simplicity) we would pick a fresh unary constant symbol $U$ and consider the expansions of $A$ and $B$ by interpreting $U$ as $\{a_1,...,a_k\}$ and $\{b_1,...,b_k\}$ respectively.