$\newcommand\C{\mathscr C}\newcommand\de{\delta}$A sufficient condition is that $E$ be compact. Indeed, since $Q$ has a separable support, without loss of generality $F$ is separable.
So, for each natural $n$ there is an (at most) countable set $\C_n$ of nonempty pairwise disjoint Borel subsets of $F$ of diameter $\le1/n$ such that $\bigcup\C_n=F$. For each $C\in\C_n$, take any $y_{n,C}\in C$ and let
$$Q_n:=\sum_{C\in\C_n}Q(C)\de_{y_{n,C}},$$
where $\de_y$ is the Dirac probability measure supported on the set $\{y\}$. Then $Q_n\to Q$ (weakly).

Next, for each $n$ and each $C\in\C_n$, take any $x_{n,C}$ such that $f(x_{n,C})=y_{n,C}$, and let
$$P_n:=\sum_{C\in\C_n}Q(C)\de_{x_{n,C}}.$$
Then $Q_n=P_nf^{-1}$, the pushforward measure of $P_n$ by $f$.

By the compactness of $E$, passing to a subsequence if needed, without loss of generality $P_n\to P$ for some probability measure $P$ on $E$. Since $f$ is continuous, it follows that $Q_n=P_nf^{-1}\to Pf^{-1}$. But $Q_n\to Q$. So, $Pf^{-1}=Q$. That is, $Q$ is the pushforward measure of $P$ by $f$, as desired.

concentratedon $f(E)$ in the sense defined by Bourbaki (the complement is locally negligible). See Intégration, Chap. V, Exercise 11 p. 125. This result I think goes back to Heinz Bauer and is further discussed in Bogachev, Measure Theory, vol. 2, p. 458. So for $f$ surjective as you assume the answer is yes. $\endgroup$