Well, if $\mathcal{A}$ is a small $k$-linear abelian category, then the embedding is given by the following:

First we put $\mathcal{A}$ inside $\mathcal{L}(\mathcal{A},\operatorname{Ab})$, the category of left exact additive functors from $\mathcal{A}$ to the category of abelian groups $\operatorname{Ab}$, by considering the contravariant Yoneda embedding $\mathcal{Y} : \mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})$ which sends $A$ to $\operatorname{Hom}_{\mathcal{A}}(A,{-})$. Since $\mathcal{A}$ is $k$-linear, we may show that $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is also $k$-linear and that $\mathcal{Y}$ is a $k$-linear functor. ($\mathcal{Y}$ is also exact.)

Now, $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is a complete abelian $k$-linear category possessing an injective cogenerator. Then we apply the duality functor $D$ in $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ and we obtain a covariant (exact) $k$-linear embedding $D \mathcal{Y} :\mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$.

Finally, we know that $\mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$ is a cocomplete abelian category possesing a projective generator $P$, and we take a certain coproduct of copies of $P$, obtaining an object $Q$. Then we take the ring $R = \operatorname{End}(Q)$, which is a $k$-algebra and we consider the exact embedding $T : \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op} \longrightarrow {\operatorname{Mod}}R$ defined by $T(X) = \operatorname{Hom}(Q,X)$, which is also $k$-linear.

Therefore, the embedding of $\mathcal{A}$ into ${\operatorname{Mod}}R$ is given by $TD \mathcal{Y} : \mathcal{A} \longrightarrow {\operatorname{Mod}}R$ and it is a $k$-linear functor.

Remarks: I took Mitchell's book "Theory of Categories" (MSN) as a reference for this answer.