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Here is just some sanity check:

We may as well work on the local case. Suppose $R=k$ local and $S=R/m$, $m$ is the maximal ideal of $R$. I will also assume $L,M$ finitely generated. Then the RHS LHS is $S^{\mu(Hom_R(L,M))}$ while the LHS RHS is $S^{\mu(L)\mu(M)}$. So if your map is an isomorphism, one must have:

$${\mu(Hom_R(L,M))} = {\mu(L)\mu(M)} \ \ \ (*)$$ Here $\mu(L)$ is the number of generators of $L$. This rarely happens unless $L$ is free. If $L$ is not, even freeness of $M$ is not enough. For example, if $M=R$ and $ann_R(L)$ contains a non-zerodivisor on $R$ (e.g, if $R$ is a domain and $L$ any torsion module), then the LHS of $(*)$ is $0$, while the RHS is $\mu(L)$.

In summary, together with the comments: if $R\otimes_kS$ is not flat over $R$, then I think $L$ must be projective for this to be true in any reasonable generality. May be your situation is more specific, if so can you tell us what you want to be true ?

EDIT: in fact, the above analysis suggests the following class of counter examples: Let $k=R$, $L = R/(x)$ where $x$ is $R$-regular and $M=R$. Then the LHS of your original map is $0$, while the LHS RHS is $Hom_S(S/(x), S) \cong 0:_{S} x$. If $x$ is not $S$-regular then the RHS is not $0$.

2 added 296 characters in body

Here is just some sanity check:

We may as well work on the local case. Suppose $R=k$ local and $S=R/m$, $m$ is the maximal ideal of $R$. I will also assume $L,M$ finitely generated. Then the RHS is $S^{\mu(Hom_R(L,M))}$ while the LHS is $S^{\mu(L)\mu(M)}$. So if your map is an isomorphism, one must have:

$${\mu(Hom_R(L,M))} = {\mu(L)\mu(M)} \ \ \ (*)$$ Here $\mu(L)$ is the number of generators of $L$. This rarely happens unless $L$ is free. If $L$ is not, even freeness of $M$ is not enough. For example, if $M=R$ and $ann_R(L)$ contains a non-zerodivisor on $R$ (e.g, if $R$ is a domain and $L$ any torsion module), then the LHS of $(*)$ is $0$, while the RHS is $\mu(L)$.

In summary, together with the comments: if $R\otimes_kS$ is not flat over $R$, then I think $L$ must be projective for this to be true in any reasonable generality. May be your situation is more specific, if so can you tell us what you want to be true ?

EDIT: in fact, the above analysis suggests the following class of counter examples: Let $k=R$, $L = R/(x)$ where $x$ is $R$-regular and $M=R$. Then the LHS of your original map is $0$, while the LHS is $Hom_S(S/(x), S) \cong 0:_{S} x$. If $x$ is not $S$-regular then the RHS is not $0$.

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Here is just some sanity check:

We may as well work on the local case. Suppose $R=k$ local and $S=R/m$, $m$ is the maximal ideal of $R$. I will also assume $L,M$ finitely generated. Then the RHS is $S^{\mu(Hom_R(L,M))}$ while the LHS is $S^{\mu(L)\mu(M)}$. So if your map is an isomorphism, one must have:

$${\mu(Hom_R(L,M))} = {\mu(L)\mu(M)} \ \ \ (*)$$ Here $\mu(L)$ is the number of generators of $L$. This rarely happens unless $L$ is free. If $L$ is not, even freeness of $M$ is not enough. For example, if $M=R$ and $ann_R(L)$ contains a non-zerodivisor on $R$ (e.g, if $R$ is a domain and $L$ any torsion module), then the LHS of $(*)$ is $0$, while the RHS is $\mu(L)$.

In summary, together with the comments: if $R\otimes_kS$ is not flat over $R$, then I think $L$ must be projective for this to be true in any reasonable generality. May be your situation is more specific, if so can you tell us what you want to be true ?