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) with maximal ideal $p$, 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 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 $K^n \oplus R^m$. If $M$ embeds into $N$, $\text{rank}(M) = \text{N}$ and $p\text{-rank}(M) = p\text{-rank}(N)$, them $M$ is quasi-isomorphic to $N$ [1, Proposition 1.3][2, Lemma 1].($M$ and $N$ are quasi-isomorphic if $M$ embeds into $N$ and $M/N$ is a torsion module bounded by a power of $p$).
The focus of [2] is the class of purely indecomposable modules (pi-modules), i.e., torsion-free indecomposable modules $M$ of finite rank with $p\text{-rank}(M) = 1$, or equivalently indecomposable pure $R$-submodules of $R^{\ast}$ of finite rank. For instance, it is shown in [2, Proposition 1], that the set of isomorphism classes of pi-modules is a partially ordered with the ascending chain condition, but not the descending one, and $R$ is its smallest element. Theorem 1 of [2] shows that the class of pi-modules is closed under direct summands and exhibit numerical invariants that determine an isomorphism class.
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.