Revision 71517ba5-9b3c-4e98-98f1-9f6ddced9fbb

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 the remains below and the enlightening counter-example of Wilberd van der Kallen.)

Still, the situation is as tamed as it can be in the case of a **complete** discrete valuation ring, see YCor's answer and his very concise proof of Joseph Rotman's theorem [1, Theorem 3].

The meaningful keywords are **[torsion-free modules](https://en.wikipedia.org/wiki/Torsion-free_module) 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)](https://en.wikipedia.org/wiki/Discrete_valuation_ring) 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 [2, Corollary 1.7]. We can also read into a representative matrix whether a module has a direct factor which is [divisible](http://mathworld.wolfram.com/DivisibleModule.html) 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 [4].

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 $N$ [2, Proposition 1.3][4, 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 [4] 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 [4, 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 [4] shows that the class of pi-modules is closed under direct summands and exhibit numerical invariants that determine an isomorphism class. 

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**Edit**: Here is a record of my first **very wrong** answer and some related developments, aggregating essentially YCor and tf_'s comments.


>> **Luc's 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 $m + k \le n$.

But the following is proved in YCor's answer, very concisely and with the exclusive use of Matlis duality.

>> **YCor's Claim (true!)**. Let $R$ be **complete** 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 $m + k \le n$.

This result turns out to be known. Indeed, it follows immediately from:

>> **Rotman's Theorem [1, Theorem 3]**. A reduced torsion-free module of finite rank over a complete DVR is free.

A module over a principal ideal ring is said to be *reduced* if its divisible submodule is $\{0\}$.

Rotman's proof is also short but relies on the Kulikov's existence of basic submodules. 

Eventually, let us note that YCor's Claim also settles:

>> **Claim 1.** Let $R$ be a complete DVR and let $K$ be its fraction field. Then $\text{Ext}_R^1(K, R) = 0$.

Claim 1, when proved by other means, can be used to prove Rotman's theorem by induction on the rank. As mentioned by tj_ in the comment, even more is true.

>> **Claim 2.** Let $R$ be a DVR with maximal ideal $p$ and let $K$ be its fraction field. Then $\text{Ext}^1(K, R) = K^{\ast}/K$ where $K^{\ast}$ is the $p$-adic completion of $K$.

Claim 2 can be used to prove the converse of Rotmans' theorem.

>> **Converse of Rotman's Theorem**. Let $R$ be a DVR with maximal ideal $p$. If every reduced torsion-free module of finite rank over $R$ is free, then $R$ is $p$-adically complete.

The proof of Claim 2 relies essentially on the computations 
$$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}.$$ 
The latter can be proved for any Noetherian local ring $R$ ideal $K/R$ is replaces by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this [MO computation](https://mathoverflow.net/questions/300453/what-is-mathrmhom-mathbbq-mathbbzp-infty/300461#300461) by YCor.

>> *Proof of Claim 2.* 
If $R$ is any domain, then $K$ is its injective hull. If $R$ is any Dedekind domain, then $K/R$ is an injective $R$ module. (If $(R, p, k)$ is a DVR with maximal ideal $p$ and residue field $k$, then $K/R$ is moreover the injective hull of $k$.)
Thus for any Dedekind domain $R$, the exact sequence $Q$
$$ 0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0$$
is an injective resolution of $R$. 
Therefore $$\text{Ext}_R^1(K, R) = H_{-1}(\text{Hom}_R(K, Q)) = \text{Hom}_R(K, K/R)/\text{Hom}_R(K, K)$$ where we have identified $\text{Hom}_R(K, K) \simeq K$ with its image in $\text{Hom}_R(K, K/R)$. From now on, the ring $R$ is assumed to be a DVR.
Let $G = \text{Hom}_R(K, K/R)$ and let $G_n \subseteq G$
the $R$-submodule of $G$ consisting of the homomorphisms which vanishes on $p^nR$. Then $G_n \simeq \text{Hom}_R(K/p^nR, K/R) \simeq \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$, the latter isomorphism being given by [3, Theorem 18.6.iv]. Since we have $G = \bigcup_{n \ge 0} G_n$ and $pG_{n + 1} = G_n$, we deduce that $G\simeq K^{\ast}$.


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[1] J. Rotman, ["A note on completion of modules"](http://www.ams.org/journals/proc/1960-011-03/S0002-9939-1960-0114813-1/S0002-9939-1960-0114813-1.pdf), 1960.  

[2] D. Arnold, "A duality Torsion-free modules of finite rank over a discrete valuation ring", 1969.  

[3] H. Matsumura, "Commutative Ring Theory", 1986.  

[4] D. Arnold, M. Dugas and K. Rangaswamy, "Torsion-free modules of finite rank over a discrete valuation ring", 2004.