Passing to the spanned subspace, it is enough to consider those submodule that span $K^n$ as a $K$-module. Then up to composition by some element of $\mathrm{GL}_n(K)$, we can suppose that the submodule $V$ contains $R^n$.
So this reduces to classifying submodules of $(K/R)^n$. This is an artinian module, actually a module over $\hat{R}$, the completion of $R$. Write $S=K/R$. Matlis duality $\mathrm{Hom}(-,S)$ yields a natural correspondence between $\hat{R}$-submodules of $S^n$ and $\hat{R}$-submodules of $\hat{R}^n$. In particular, modulo composition by an element of $\mathrm{GL}_n(\hat{R})$, every submodule of $S^n$ can be written as $S^k\times F$, where $F=\prod_{i=k+1}^nF_i\subset S^{n-k}$ has finite length. Actually, we can suppose $F=0$, composing beforehand by another element of $\mathrm{GL}_n(K)$.
In particular, Luc's initial claim is correct when $R$ is complete (but fails as soon as it's not complete and $n\ge 2$).
