Prove: $ T ◦ S $ is surjective iff $ $ T is surjective and $ ImS + KerT = V $

I think it might be the easiest to verify the equality by direct computation via elements instead of dimension (the latter approach is actually a bit problematic if the vector spaces are no longer finite-dimensional)
For $\Rightarrow$: Let $v\in V$ be arbitrary. We wish to express it as a sum of elements in $\ker T$ and $\operatorname{im} S$. For this notice that since $TS$ is surjective there is some $u\in U$ with $T(v)=T(S(u))$, so we have $v-S(u)\in\ker T$. Now $$v=(v-S(u))+S(u)$$ is the desired decomposition.
The reverse implication is similar. Take $w\in W$. By assumption there is some $v\in V$ with $T(v)=w$. Furthermore we have $v=x+y$ with $x\in\ker T,y\in \operatorname{im} S$, i.e. $x=S(u)$ for some $u\in U$. Check that $T(S(u))=w$.

(I left out some details in the answer, which you might need to fill in)

For the "$\Leftarrow$" part: Let $T$ be surjective and let's assume $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$. Let $w \in W$. We need to find $u \in U$ such that $T(S(u))=w$.

Since T is surjective, $\exists v \in V$ such that $w=T(v)$. Since $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$, there are some $v_1 \in \operatorname{Ker}(T), v_2 \in \operatorname{Im}(S)$ such that $v=v_1+v_2$. Can you continue from here?