When we have a $G$-modules exact sequence $0\rightarrow A\stackrel{f}\rightarrow B\stackrel{h}\rightarrow C\rightarrow 0$ we have a $G$-modules exact sequence $0\rightarrow A^G\stackrel{f}\rightarrow B^G\rightarrow C^G\rightarrow 0$ when $|G|<\infty$?
Here $M^G:=\{ m\in M|\ g\cdot m=m$ for all $g\in G\}$
Proof : I have a difficulty to prove the surjectivity of $h$. For $c\in C^G$ there exists $b\in B$, $h(b)=c$. If $g\cdot b\neq b$ for some $g\in G$, define $b_0:= \sum_{g\in G} g\cdot b$ Then we have $$ h(b_0)=|G| c $$ Here $g\cdot b_0=b_0$
But I have no idea further. How can we finish ?
As I said in the comments your claim is false. Here is an explicit counterexample:
Let $G = \{ \pm 1 \}$, and let $B = \mathbb{Z}/4$, $C = \mathbb{Z}/2$ be $G$-modules in the obvious way. Then $h : B \to C$ with $h(1) = 1$ is an epimorphism of $G$-modules, but $h^G : B^G \to C^G$ is the zero map.