All the question is in the title: is the equivalency $$(\forall (x,y) \in F\times G, \phi(x) \lor \psi(y) ) \Leftrightarrow (\forall x\in F, \phi(x)) \lor (\forall y\in G, \psi(y))$$ true (where $\phi(x)$ (resp. $\psi(y)$) is a formula where $y$ (resp. $x$) doesn't appear)?
I would like say yes, but I'm not sure: do we have $$(\forall x\in F (\phi(x)\lor \psi(y))) \Leftrightarrow ((\forall x\in F, \phi(x))\lor \psi(y))$$ If yes, I could say:
$$\begin{aligned} (\forall (x,y) \in F\times G, \phi(x) \lor \psi(y) )&\Leftrightarrow (\forall y\in F(\forall x\in G (\phi(x)\lor \psi(y))) \\ &\Leftrightarrow (\forall y\in F((\forall x\in G, \phi(x))\lor \psi(y)) \\&\Leftrightarrow (\forall x\in F, \phi(x)) \lor (\forall y\in G, \psi(y)) \end{aligned}$$
Yes, as long as $x$ is not free in $\psi_y$, then $\psi_y$ is constant with respect to $x$ - its validity is not affected by varying $x$ in any way.
$$\forall x (x\in F\to \phi_x \vee \psi_y)\quad\text{ iff }\quad\forall x (x\in F\to \phi_x) \vee \psi_y$$
$$``\text{any $x$ in $F$ satisfies either $\phi_x$ or $\psi_y$" iff $``$either $\psi_y$ is true or every $x$ in $F$ satisfies $\phi_x$"}$$
Slightly more formally:
Assuming $\psi_y$, then $\psi_y, \forall x (x\in F\to \phi_x \vee \psi_y) \vdash \forall x(x\in F\to \psi_y) \vdash \psi_y$
Assuming $\neg\psi_y$, $\neg \psi_y, \forall x (x\in F\to \phi_x \vee \psi_y) \vdash \forall x(x\in F\to \phi_x)$
Therefore, $\forall x (x\in F\to \phi_x \vee \psi_y) \vdash \forall x(x\in F\to \psi_x)\vee \psi_y$
And you can prove the converse.
PS: we should also ensure $F$ is not empty.