The matrix category is fully faithful in modules and, with chosen bases, equivalent to finite free modules

Define $K$ on objects by $K(n)=R^n$ and on morphisms by matrix multiplication on column vectors. Commutativity of $R$ makes these maps left $R$-linear, and associativity and unitality of matrix multiplication make $K$ a functor into the module category of [L3].

Fix $n,m$. An $m\times n$ matrix is exactly a choice of image in $R^m$ for each standard basis vector of $R^n$, namely its columns. By [L2], such choices are in bijection with $R$-linear maps $R^n\to R^m$. Therefore $K$ is fully faithful.

Let $M$ be finitely generated and free. Choose a basis $B$ and a finite generating set $S$. Each element of $S$ is a linear combination of finitely many elements of $B$, so the union $B_0$ of these finite supports is finite. Since $S$ generates $M$, the set $B_0$ also generates $M$; if $b\in B\setminus B_0$, expressing $b$ in the span of $B_0$ would contradict uniqueness of coordinates in the basis $B$. Thus $B=B_0$ is finite. After enumerating it with $n$ elements, [L2] identifies $M$ with $R^n$. Conversely, each $R^n$ is finitely generated and free, so the essential image of $K$ is exactly the full subcategory of finitely generated free modules. The supplied pairs $(n_M,R^{n_M}\cong M)$ make this essential surjectivity split.

By steps 1.2 and 1.3, the functor $K$ is fully faithful and split essentially surjective onto that full subcategory, so [L4] gives an equivalence. The source matrix category is additive by [L5].