How to prove Hom$_{k}(M,N)^{G}\cong$Hom$_{kG}(M,N)$ $M,N$ are $kG$-module, $G$ group and Hom$_{k}(M,N)^{G}$ is the invariants of Hom$_{k}(M,N)$ which I believe is $\{ f\in \text{Hom}_k (M,N) | gf=g \forall g \in G \}$?
And the isomorphism here is isomorphism between $kG$-modules, $k$-modules or abelian groups?

This boils down to understanding the bimodule structure on $\hom_k(M,N)$. Given such $k$-linear map $\varphi:M\to N$, we define $(g\varphi )(x)=g \varphi(x)$ and $(\varphi g )(x)=\varphi(gx)$. The $G$-invariants are thus the $k$-linear maps that satisfy $\varphi(gx)=g\varphi(x)$ for every $x\in M$ and $g\in G$, i.e. the $kG$-linear maps. 
More generally, suppose $H$ is a Hopf algebra over a field $k$ and $M,N$ are $H$-modules. Then $\hom_k(M,N)$ admits an $H$-module structure as follows. Take a $k$-linear map $\varphi:M\to N$, and $h\in H$. Apply $\Delta(h)=\sum h_1\otimes h_2$, then apply $1\otimes S$ to get $\sum h_1\otimes S(h_2)$, finally apply $1\otimes f$ and $\mu$, to get $$(h\cdot f)(a)=\sum h_1 f(S(h_2)a)$$
In the case $H=kG$ we have $S(g)=g^{-1}$ and $\Delta(g)=g\otimes g$ so that $$(g\cdot f )(a)=g f(g^{-1}a)$$
and the invariant morphisms are precisely those that are $G$-linear. 
