From Wikipedia:
[O]ne instance of the schema is included for each formula $\phi$ in the language of set theory with free variables among $x, y, A, w_1,\ldots, w_n$; but $B$ not free in $φ$. In the formal language of set theory, the axiom schema is: $$ \forall A \forall w_1, \dots, w_n \; \Big(\big[\forall x \in A \; \exists! y \; \phi(x, y, w_1, \dots, w_n, A)\big] \implies \exists B \; \forall y \; \big[y \in B \iff \exists x \in A \; \phi(x, y, w_1, \dots, w_n, A)\big]\Big) $$
[...] For clarity, in the case of no variables $w_i$, this simplifies to:
$$ \forall A \; \Big(\big[\forall x \in A \; \exists! y \; \phi(x, y, A)\big] \implies \exists B \; \forall y \; \big[y \in B \iff \exists x \in A \; \phi(x, y, A)\big]\Big) $$
I'm struggling to understand the usefulness of allowing $\phi$ to have other free variables besides $x$ and $y$. Even in the simplified version, I do not see the merit of including $A$ as a free variable in $\phi$.
All the applications of the axiom that come to mind are like so:
Let $A\subseteq \mathbb{Z}$, and $\phi(x,y)$ represent the equation $y = 2x$. It holds that $$\forall x \exists ! y (y = 2x)$$ and thus the axiom gives that there is some $B$ for which $$\forall y\big[y \in B \iff \exists x \in A (y = 2x)\big]\Big).$$ or, more formally, $$\forall y\big[y \in B \iff \exists x (x \in A \land y = 2x)\big]\Big)$$ e.g. $A = \{1, 3, 5\}$, then $B = \{2, 6, 10\}$.
Why do we allow $\phi$ to have free variables $w_1,\ldots,w_n,A$?
