In A.G. Hamilton's Logic for mathematicians, eight axioms of ZF are given: EXT, NULL, PAIR, UNION, POW, REP, INF and REG.
The Axiom Scheme of Replacement is formulated like this: $$ (\forall x_1)(\exists! x_2)\mathscr{A}(x_1,x_2)\to(\forall x_3)(\exists x_4)(\forall x_5)(x_5\in x_4\leftrightarrow(\exists x_6)(x_6\in x_3\wedge\mathscr{A}(x_6,x_5))) $$
My question is:
Since SEP is not taken as an axiom, shouldn't REP be formulated $$ (\forall x_1)((\forall x_2)(x_2\in x_1\to(\exists! x_3)\mathscr{A}(x_2,x_3))\to(\exists x_4)(\forall x_5)(x_5\in x_4\leftrightarrow(\exists x_6)(x_6\in x_1\wedge\mathscr{A}(x_6,x_5)))) $$
If I understood your reformulation correctly, you're trying to capture the idea that we must first have a set in order to be able of specifying (or separating) a further set by a property (if that's not the idea, ignore my answer). That "patch", however, is not necessary: the consequent of the axiom already specify that the "values" of the function-like expression must belong to a set in order for there to be a further set containing their "images". Notice that the existential quantifier $\exists x_6$ is dependent on the previous quantifier $\forall x_3$; this means that the consequent states: for every set $x_3$, there is a further set $x_4$ such that for every set $x_5$, $x_5$ is an element of $x_4$ iff there is an element $x_6$ of $x_3$ such that $\mathscr{A}(x_6, x_5)$.
As for Suppes, note that he's working in a set-theory with urelements (objects which are not sets). That's why he includes a free variable (not bound, as in your case) for sets in his formulation of the axiom.