I have a statement of the Tarski-Vaught test as follows.
Let $\mathscr{M}$ be an $\mathscr{L}$-structure and let A $\subseteq |\mathscr{M}|$. Then the following are equivalent:
(i) A is the universe of an elementary substructure of $\mathscr{M}$. Since on each subset of $\mathscr{M}$ there can only live at most one substructure of $\mathscr{M}$, this property is referred to as "A is a substructure of $\mathscr{M}$".
(ii) For every $\mathscr{L}$-formula $\phi(x, \bar{y})$ and all $\bar{a} \in A^{\bar{y}}$, if $\mathscr{M} \vDash \exists x \phi(x, \bar{a})$, then there is some $b \in A$ with $\mathscr{M} \vDash \phi(b, \bar{a})$.
I'm unsure about part (i) - why is it that on each subset of $\mathscr{M}$ there can only live at most one substructure of $\mathscr{M}$?
That's by definition of a substructure. Given a subset $A$ which contains all constant terms, you can make it into an $L$-structure by restricting all functions and relations from $M$ to $A$ (i.e. by removing all tuples which have elements not in $A$).
This $L$-structure is considered a substructure if your functions didn't lose any elements. I.e. if $f$ is an $n$-ary function and $M\vDash f(a_1,\ldots,a_n)\doteq b$ for $a_1,\ldots,a_n \in A$ then $b \in A$.
If the former holds then $A$ is a substructure (with the particular interpretation for the symbols of $L$ described above), and else it is not.
Now Vaught test gives you a criterion to decide whether $A$ (considered as the structure described above) is also elementary, i.e. whether the theories of $A$ and $M$ coincide.