Indeed, it seems that the situation gets nicer, but certainly not as nice as what I depicted in my first and very flawed answer (see its remains below and the enlightening counter-example of Wilberd van der Kallen).
The meaningful keywords are torsion-free modules of finite rank over a discrete valuation ring. Note that if $R$ is any domain and $K$ its fraction field, then any given torsion-free $R$-module $M$ of finite rank $n$ embeds into $K^n = M \otimes_R K$. Conversely, any $R$-submodule of $K^n$ is torsion-free of finite rank.
If $(R, p)$ is a discrete valuation ring (DVR), then a torsion-free $R$-module of finite rank $n$ can be represented by an $n \times n$ unipotent matrix with coefficients in $R^{\ast}$, the $p$-adic completion of $R$. The $R$-isomorphism and quasi-isomorphism classes of the torsion-free $R$-modules of finite rank can be described via three different conditions on such matrices [1, Corollary 1.7]. We can also read into a representative matrix whether a module has a direct factor which is divisible or free [2, Corollary 1.8]. This should help build many examples of indecomposable modules with no such factors such as Wilberd van der Kallen's. Other useful results can be found in [2].
If $M$ and $N$ are two isomorphic torsion-free $R$-modules of finite rank, then such modules have the same rank and the same $p\text{-rank}$, where $p\text{-rank}(M) = \dim_k(M/pM)$ and $k = R/p$. But these two invariants are not complete and most torsion-free $R$-modules of finite rank are not of the form $K^n \oplus R^m$.
Edit: Here is a record of my first very wrong answer.
Claim (wrong!). Let $R$ be DVR and let $K$ be its fraction field. An $R$-submodule of $K^n$ is isomorphic to an $R$-module the form $K^m \oplus R^k$ for some non-negative integers $m$ and $k$ such that $m + k \le n$.
[1] D. Arnold, "A duality Torsion-free modules of finite rank over a discrete valuation ring", 1969.
[2] D. Arnold, M. Dugas and K. Rangaswamy, "Torsion-free modules of finite rank over a discrete valuation ring", 2002.