Not sure if this is what you are looking for, but certainly this is a necessary condition for your condition to hold.
Let $S$ be a central subring of a (not necessarily commutative) ring $R$, with $R$ free and finite dimensional as a module over $S$. Your condition is that $R$ is isomorphic to ${\rm Hom}_S(R,S)$ as left $R$ modules. Equivalently, there exists a map $\epsilon\colon R\to S$ such that every $S$-linear homomorphism $R \to S$ may be written in the form $\epsilon(\_a)$ for a unique $a\in R$.
A necessary condition for such an $\epsilon$ to exist is that finitely generated projective $R$-modules are injective relative to $S$. That is given an $R$-linear map of left $R$ modules $f\colon A \to M$ such that $f$ has a left inverse as a map of $S$ modules, any $R$-linear map $h\colon A \to P$ (for $P$ a finitely generated projective module) may be extended to an $R$-linear map $M \to P$.
$$
A\stackrel f\to M
$$
$$h\downarrow \,\,\,\swarrow\quad$$
$$P\quad\quad$$
Proof: Suppose $\epsilon$ exists as above. It is sufficient to consider the case $P=R$, as the property of being relatively injective extends in an obvious way to (finite) direct sums and summands.
Given $m\in M$ we have an element of ${\rm Hom}_S(R,S)$ given by $$\lambda\mapsto \epsilon(hg(\lambda m))$$ where $g$ is the $S$-linear left inverse to $f$.
Thus we have $\hat h(m)\in R$ such that $$\epsilon(hg(\lambda m))=\epsilon(\lambda \hat h(m),$$
for all $\lambda\in R$. Then $\hat h$ is $R$-linear as for all $\lambda\in R$ we have $$\epsilon (\lambda \hat h(\mu m))=\epsilon(hg(\lambda\mu m))=\epsilon(\lambda\mu\hat h(m)).$$
Finally we note that $\hat hf=h$: $$\epsilon(\lambda\hat hf(a))=\epsilon(hg(\lambda f(a)))=\epsilon(hgf(\lambda a))=\epsilon(h(\lambda a))=\epsilon(\lambda h(a)),$$ for all $\lambda \in R, \,\, a \in A$.
Edit:
I will first address the general form of the Tensor-Hom adjunction.
In general, the Tensor-Hom adjunction states that if $X$ is an $(R, S)$-bimodule, $Y$ an $(A, R)$-bimodule and $Z$ a $(B, S)$-bimodule then $$\mathrm{Hom}_S(Y \otimes_R X, Z) = \mathrm{Hom}_R(Y, \mathrm{Hom}_S(X, Z))$$ as $(B, A)$-bimodules.
The proof is as follows: Since $X$ is a right $S$-module, we have a right $S$-module structure on $Y \otimes_R X$. Thus $\mathrm{Hom}_S(Y \otimes_R X, Z)$ makes sense. Now, since $Y$ is a left $A$-module, we have a left $A$-module structure on $Y \otimes_R X$, so we can define a right $A$-module structure on $\mathrm{Hom}_S(Y \otimes_R X, Z)$ by sending $f : Y \otimes_R X \to Z$ to the homomorphism $(fa)(p) = f(ap)$ (verify this). Since $Z$ is a left $B$-module, we can define a left $B$-module structure by $(bf)(p) = bf(p)$. Now $((bf)a)p = (bf)(ap) = bf(ap) = b((fa)p) = (b(fa))(p)$, so we have a $(B, A)$-bimodule.
Similarly, we have left $R$-module structure on $X$, so we can define a right $R$-module structure on $\mathrm{Hom}_S(X, Z)$ as follows: for $f : X \to Z$ we define $fs$ by $(fs)(x) = f(sx)$. Thus $\mathrm{Hom}_R(Y, \mathrm{Hom}_S(X, Z))$ makes sense. Now since $Z$ is a left $B$-module,we can define a left $B$-module structure on $\mathrm{Hom}_S(X, Z)$ by sending $f: X \to Z$ to the morphism $(bf)(x) = bf(x)$. Similarly we get a left $B$-module structure on $\mathrm{Hom}_R(Y, \mathrm{Hom}_S(X, Z))$. The right $A$-module structure is given by sending $f : Y \to \mathrm{Hom}_S(X, Z)$ to $(fa)(y) = f(ay)$. You can verify these structure are compatible and you get a $(B, A)$-bimodule.
We now show they are isomorphic as $(B, A)$-bimodules. Let $f : Y \otimes_R X \to Z$. We define $$\varphi : \mathrm{Hom}_S(Y \otimes_R X, Z) = \mathrm{Hom}_R(Y, \mathrm{Hom}_S(X, Z))$$ by $\varphi(f)(y)(x) = f(y \otimes x)$. You know how to show additivity, so I will only show linearity.
Let $b \in B$, then $$\begin{aligned}
\varphi(bf)(y)(x) &= (bf)(y \otimes x) \\
&= bf(y \otimes x) \\
&= b (\varphi(f)(y))(x) \\
&= (b(\varphi(f)(y)))(x) \\
&= ((b\varphi(f))(y))(x) \\
&= (b\varphi(f))(y)(x),
\end{aligned}$$
by filling in definitions and using the appropriate module structures (parentheses are hopefully clear enough where necessary).
Similarly, for $a \in A$ we have $$\begin{aligned}
\varphi(fa)(y)(x) &= (fa)(y \otimes x) \\
&= f(a(y \otimes x)) \\
&= f(ay \otimes x) \\
&= (\varphi(f)(ay))(x) \\
&= (\varphi(f)a)(y)(x)
\end{aligned}.
$$
For an inverse, let $g: Y \to \mathrm{Hom}_S(X, Z)$, then define $\psi(g) : Y \times X \to Z$ by $\psi(g)(y, x) = g(y)(x)$, check the usual conditions to get a map from $Y \otimes_R X \to Z$ and show that it is an inverse.
Addressing your question:
Let $P$ be a $(R, S)$-bimodule, $N$ a left $S$-module and $M$ a right $R$-module.
Then $P \otimes_S N$ has a structure of a left $R$ module, while $M$ has the structure of a right $R$-module.
Now, a module homomorphism $f : P \otimes_S N \to M$ must satisfy $f(rp) = f(p)r$, but then $f(p)rr' = f((rr')p) = f(r(r'p)) = f(r'p)r = f(p)r'r$, so the module structures on $P, N, M$ don't seem like very natural choices, unless the rings are commutative. I believe the following will work under that assumption.
I am not entirely sure if this is what you're asking, but I will show that $$S(\phi \cdot s) = S(\phi) \cdot s.$$
You haven't mentioned it explicitly in your post, but the module structure on $\mathrm{Hom}_S(N, \mathrm{Hom}_R(P, M))$ is given by sending the morphism of $S$-modules $f : N \to \mathrm{Hom}_R(P, M)$ to the morphism $f \cdot s$ sending $n \in N$ to $f(n \cdot s) = f(n) \cdot s$. Here $f(n) \in \mathrm{Hom}_R(P, M)$ and the action is given by the one in your post.
We show first that $S(\phi \cdot s)(n) = (S(\phi) \cdot s)(n)$: note that for any $p \in P$ we have that $$\begin{aligned}
S(\phi \cdot s)(n)(p) &= (\phi \cdot s)(p \otimes n) \\
&= \phi (ps \otimes n) \\
&= S(\phi)(n)(ps) \\
&= (S(\phi)(n) \cdot s)(p) \\
&= S(\phi)(n \cdot s)(p) \\
&= (S(\phi) \cdot s)(n)(p).
\end{aligned}$$
Here the first 3 lines is just unwinding definitions, the fourth line uses the module structure on $\mathrm{Hom}_R(P, M)$, the fifth line the fact $S(\phi)$ is a homomorphism of $S$-modules, and the sixth line the module structure of $\mathrm{Hom}_S(N, \mathrm{Hom}_R(P, M))$.
Best Answer
You are right in saying that tensoring $k[G]$-modules like you do here is different than tensoring modules over rings, because in general if $A$ is a ring and $M$ and $N$ are $A$-modules, $M\otimes N$ is not an $A$-module, but simply an $(A\otimes A)$-module. The miracle here is that $k[G]$ is much more than a ring, it is a Hopf algebra, and in particular it has a comultiplication $k[G]\to k[G]\otimes k[G]$ (given by $g\mapsto g\otimes g$) which allows us to turn any $(k[G]\otimes k[G])$-module into a $k[G]$-module.
Now about your question. The first step is to see that at least the isomorphism holds as $k$-vector spaces, which I think you know how to do: $$ \operatorname{Hom}_R(R\otimes_k V, R\otimes_k W)\simeq \operatorname{Hom}_k(V, R\otimes_k W)\simeq R\otimes_k W\otimes_k V^* $$ with the isomorphism $\Phi: R\otimes_k W\otimes_k V^*\to \operatorname{Hom}_R(R\otimes_k V, R\otimes_k W)$ given by $$ r\otimes w\otimes f \mapsto \left( a\otimes v\mapsto f(v)\cdot ar\otimes w\right).$$
The fact that it is an isomorphism hinges on the fact that $V$ is finite-dimensional.
So we just have to check that this respects the $G$-actions. Now: $$\Phi(gr\otimes gw\otimes gf)(a\otimes v) = f(g^{-1}v)\cdot a(gr)\otimes (gw)$$ and $$\begin{align*} g(\Phi(r\otimes w\otimes f))(a\otimes v) &= g\left( \Phi(r\otimes w\otimes f)(g^{-1}a\otimes g^{-1}v) \right)\\ &= g\left( f(g^{-1}v)\cdot (g^{-1}a)r\otimes w \right) \\ &= f(g^{-1}v)\cdot a(gr)\otimes gw. \end{align*}$$
So indeed this is an isomorphism of $k[G]$-modules. Note that this requires that the action of $G$ on $R$ is by algebra automorphisms, which I think is implicit in your question (so we should have $g(ab)=(ga)(gb)$ if $a,b\in R$).