$R$ a DVR with fraction field $K.$ What are the $R$-submodules of $K^n?$
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 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 [2, 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 [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 $N$, $\text{rank}(M) = \text{rank}(N)$ 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 set 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.
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 $\text{Ext}^1(K, R) = R^{\ast}/R$ where $R^{\ast}$ is the $p$-adic completion of $R$.
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 $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$$ where $K^{\ast}$ is the $p$-adic completion of $K$. The latter can be proved for any Noetherian local ring $R$ if $K/R$ is replaced 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 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 $G\simeq K^{\ast}$. It is not difficult to show that $K^{\ast}/K \simeq R^{\ast}/R$. Indeed, the kernel of the map induced by the inclusion $R^{\ast} \rightarrow K^{\ast}/K$ is $R^{\ast} \cap K = R$ while $K^{\ast} = R^{\ast} + K$ follows from the density of $R$ in $R^{\ast}$.
[1] J. Rotman, "A note on completion of modules", 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.
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 the quotient $\hat{R}$-modules 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 $n\ge 2$).
Fix a prime $p$ and let $x$ be a $p$-adic integer that is not a rational number. Now let $M$ be the group of pairs of rational numbers $(r,s)$ so that $rx-s$ is a $p$-adic integer. Then $M$ maps onto $\mathbb Q$ but it does not contain a copy of $\mathbb Q$, because that would mean that some $s/r$ would not just be a good approximation of $x$, but a perfect one.