Let $G = \prod_F F$ be the direct product of all finite groups (one per isomorphism class, obviously) and let $F_\omega$ be the free group on a countably infinite collection of generators. Obviously $G$ and $F_\omega$ both cover every finite group, and there is obvioulsy no surjective homomorphism $F_\omega \to G$ because $G$ is uncountable. We claim there is no nontrivial homomorphism $G \to F_\omega$.

Suppose $g$ is an element of a finite group $F$ and let $A = \langle g \rangle$ be the cyclic group generated by $g$. Then $A = U \times V$ for a cyclic $2$-group $U$ and a cyclic group $V$ of odd order. It follows that we may write $g = uv$ where $u$ has $2$-power order and $v$ has odd order (and $u$ and $v$ commute). Note that $u$ is infinitely $3$-divisible and $v$ is infinitely $2$-divisible.

Now let $f : G = \prod_F F \to F_\omega$ be a homomorphism. Let $g \in G$. Applying the above to each component of $g$, we may write $g = uv$ where $u$ is infinitely $3$-divisible and $v$ is infinitely $2$-divisible. But $F_\omega$ does not contain any nontrivial infinitely $2$- or $3$-divisible elements (since it has a well-defined notion of "length"), so $f$ must kill $u$ and $v$ and thus $g$ too. This shows that $f$ is trivial.

Not a complete answer: Let $G = \prod_F F$ be the direct product of all finite groups and let $F_\omega$ be the free group on a countably infinite collection of generators. Obviously $G$ and $F_\omega$ both cover every finite group. To show that there is no group $S$ as sought in the question it suffices to show that $G$ has no countable quotient that still covers every finite group. Does it?