I'm trying to work my way through back-and-forth systems and elimination sets by reading the relevant sections in Hodges' Model Theory and I'm a bit confused by one of his lemmas (specifically, it's lemma 3.3.5). Let me start with a few definitions, which I think are clear enough (I'm assuming throughout that every structure is an $\mathcal{L}$-structure and thus that every formula is an $\mathcal{L}$-formula):
Definition 1: An unnested Enhrenfeucht-Fraïssé game of length $k$, in symbols $\mathrm{EF}_k [(\mathcal{A}, \bar{a}), (\mathcal{B}, \bar{b})]$, is a game between two players, $\exists$ and $\forall$, such that, at the $i$th move in the game, $\forall$ must choose an element from either $\mathcal{A}$ or $\mathcal{B}$; then player $\exists$ chooses an element from the other structure. $\exists$ wins the game if, at the end of the play, the players have chosen $k$-tuples $\bar{c}$ and $\bar{d}$ from $\mathcal{A}$ and $\mathcal{B}$, respectively, such that, for every unnested atomic formula $\phi$, $\mathcal{A} \models \phi(\bar{a}, \bar{c})$ iff $\mathcal{B} \models \phi(\bar{b}, \bar{d})$; otherwise, Player $\forall$ wins. If $\exists$ wins, then we write $(\mathcal{A}, \bar{a}) \sim_k (\mathcal{B}, \bar{b})$.
Let now $\mathscr{K}$ be a class of structures. For each $\mathcal{A} \in \mathscr{K}$, let $\mathrm{tup}(\mathcal{A})$ be the set of all pairs $(\mathcal{A}, \bar{a})$, with $\bar{a}$ a tuple from $\mathcal{A}$, and let $\mathrm{tup}(\mathscr{K})$ be the union of all $\mathrm{tup}(\mathcal{A})$ for $\mathcal{A} \in \mathscr{K}$.
Definition 2: By an unnested graded back-and-forth system for $\mathscr{K}$, we mean a family of equivalence relations $(E_k \; : \; k \in \omega)$ such that:
(1) if $\bar{a} \in \mathrm{tup}(\mathcal{A})$ and $\bar{b} \in \mathrm{tup}(\mathcal{B})$, then $\bar{a}E_0\bar{b}$ iff for every unnested atomic formula $\phi(\bar{x})$, $\mathcal{A} \models \phi(\bar{a})$ iff $\mathcal{B} \models \phi(\bar{b})$;
(2) if $\bar{a} \in \mathrm{tup}(\mathcal{A})$ and $\bar{b} \in \mathrm{tup}(\mathcal{B})$, then $\bar{a}E_{k+1}\bar{b}$ iff for every $c \in A$ there is $d \in B$ such that $\bar{a}cE_k\bar{b}d$.
Now, it's not difficult to see (this is Hodges' Lemma 3.3.4) that, if $\bar{a}E_k\bar{b}$, then $(\mathcal{A}, \bar{a}) \sim_k (\mathcal{B}, \bar{b})$. But then comes my question: Hodges then claims next (p. 106) that this implies that each equivalence class under $\sim_k$ is a union of equivalence classes of $E_k$. I mean, it's clear that every $E_k$ is a subset of the corresponding $\sim_k$, by this result. But how can I get the result that, if $(\mathcal{A}, \bar{a}) \sim_k (\mathcal{B}, \bar{b})$, then there is an equivalence class of $E_k$ to which both belong?
The fact that each $\sim_k$-class is a union of $E_k$-classes really does follow immediately from the fact (Lemma 3.3.5) that each $E_k$-class is contained in a $\sim_k$-class. Indeed, writing $[(A,\overline{a})]_{E_k}$ for the $E_k$-class of $(A,\overline{a})$, if $C$ is a $\sim_k$-class, then $C = \bigcup_{(A,\overline{a})\in C}[(A,\overline{a})]_{E_k}$.
You write "How can I get the result that, if $(\mathcal{A}, \bar{a}) \sim_k (\mathcal{B}, \bar{b})$, then there is an equivalence class of $E_k$ to which both belong?". What you want, together with Lemma 3.3.5, would imply that $E_k$ and $\sim_k$ are equal. Hodges doesn't claim this, and it's not true in general. Rather, $E_k$ is a finer equivalence relation than $\sim_k$.