The two questions can be answered in the positive.

**Claim 1.** Let $R$ be an infinite division ring. Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.

The proof is a straightforward generalization of the classical pigeonhole argument for finite covers of vector spaces, see the Corollary below.

If $R$ is a commutative ring with identity which has a finite residue field $k$, then any vector space over $k$ with finite dimension $ > 1$ is an $R$-module that can be covered by finitely many (proper) lines. OP's second question is therefore addressed by

**Claim 2.** Let $R$ be a commutative ring with identity and whose residue fields are all infinite. Then no $R$-module can be the union of finitely many of its proper $R$-submodules.

The class of rings which satisfy the hypotheses of Claim 2 is wide: fix finitely many fields $k_1, \dots, k_n$ and let $\{K_{\alpha}\}_{\alpha}$ be a family of fields such for every $\alpha$, the field $K_{\alpha}$ contains (up to isomorphism) one of the fields $k_i$. Then there is a zero-dimensional ring $R$ whose residue fields are the fields $K_{\alpha}$ [1].

The proof of Claim 2 relies on [3, Lemma 10]:

**Gottlieb's Lemma.** Let $R$ be a commutative ring with identity.
Let $M = M_1 \cup M_2 \cup \cdots \cup M_n$ be an $R$-module written as an union of proper submodules and suppose there is an element $x \in M_1 \setminus \bigcup_{i = 2}^n M_i$ and elements $a_1, a_2, \dots, a_r \in R$ which are pairwise distinct modulo every ideal
$(M_k: x), k \ge 2$. Then $n > r$.

If $M$ is a module over $R$, $x \in M$, and $N$ is a submodule of $M$, we denote by $(N:x)$ the ideal of $R$ consisting of the elements $r$ such that $rx \in N$.

Now we can prove the second claim.

*Proof of Claim 2.* Let $M = M_1 \cup M_2 \cup \cdots \cup M_n$ be an $R$-module written as an union of proper submodules. We can assume that $n > 2$ and $M_1 \not\subset M'$ with $M' \Doteq M_2 \cup \cdots \cup M_n$. Let $x \in M_1 \setminus M'$ and pick some $r \ge n$. Each ideal $(M_k: x)$ for $k \ge 2$ is contained in some maximal ideal $\mathfrak{m_i}$ for $i = 1, \dots, N$ and $N \le k - 1$. Since $R/\mathfrak{m}_i$ is infinite, we can find for every $i$ some elements $a_{i1}, \dots, a_{ir} \in R$ which are pairwise distinct modulo $\mathfrak{m}_i$. By the Chinese Remainder Theorem, we can find $a_1, \dots, a_r \in R$ such that $a_j \equiv a_{ij} \text{ mod } \mathfrak{m}_i$ for every $i, j$. Gottlieb's Lemma yields $n > r$, a contradiction.

**Addendum.** I realized that Claim 2 is also an immediate consequence of [2, Lemma 3].

[1] R. Gilmer, W. Heinzer, "The family of residue fields of a zero-dimensional commutative ring", 1992.

[2] M. Zandt, "A note on unions of ideals and cosets of ideals", 1995.

[3] C. Gottlieb, "Modules covered by finite unions of submodules", 1998.

Below lie the remains of my initial (awkward) answer.

This is only a long comment.

Let $R$ a be unital ring which is not necessarily commutative and let
$R^{\times}$ denote the set of left-invertible elements of $R$. We say that $R$ has property (U) if for every finite cover $R^{\times} \subset \bigcup_{i = 1}^n A_i$
of $R^{\times}$ by subsets $A_i \subset R$, there is at least
one index $j$ such that $A_j$ contains two left-invertible elements $\lambda, \mu$
satisfying $\lambda - \mu \in R^{\times}$.

**Claim 3.** Let $R$ a be unital ring with property (U). Then no left
$R$-module can be the union of finitely many of its proper left $R$-submodules.

*Proof.* We reason by contradiction, considering a left $R$-module $M$
that is covered by a finitely many proper left $R$-submodules $M_1, \dots, M_n$.
Without loss of generality, we can assume that $n > 1$, $M'
\Doteq\bigcup_{i = 1}^{n -1} M_i \not \subset M_n$ and $M_n \not\subset
M'$. Consider $z(u) = x + uy$ with $x \in M_n \setminus M'$ and $y \in
M' \setminus M_n$ and $u \in R^{\times}$. We have $z(u) \notin M_n$ since
otherwise $y$ would lie in $M_n$ too. Therefore $z(u) \in M'$ for every $u \in R^{\times}$. Set $A_i = \{ u \in R^{\times}\, \vert\, z(u) \in M_i\}$ for $i = 1, \dots, n - 1$. As $R$ has property (U), we can find $\lambda, \mu \in A_j$ for some $j \le n - 1$ such that $\lambda - \mu \in R^{\times}$. As $z(\lambda) - z(\mu) \in M_j$ we deduce that $y \in M_j$, a contradiction.

This rewording of the classical proof has an immediate consequence:

**Corollary.** Assume that $R$ is an infinite division ring or a commutative semilocal ring with identity and whose residue fields are infinite. Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.

*Proof.* If $R$ is an infinite division ring, then $R^{\times} = R \setminus \{0\}$ so that $R$ has clearly property (U) and the result follows from Claim 3. Assume now that $R$ is a unital commutative ring with finitely many maximal ideals $\mathfrak{m}_1, \dots, \mathfrak{m}_n$. Then $R^{\times} = R \setminus \bigcup_{i = 1}^{n} \mathfrak{m}_i$. Assume that $R^{\times} \subset
\bigcup_{j = 1}^N A_j$ for some subsets $A_j \subset R$. We reason by contradiction, supposing further that the difference of any pair of units in $A_j$ is not invertible for every $j$. Then $R^{\times} \subset \bigcup_{1 \le i \le n, 1 \le j \le N} (x_{ij} + \mathfrak{m}_i)$ for some elements $x_{ij} \in R$. Since $R/\mathfrak{m}_i$ is infinite for every $i$, we can find $x_i \in R$ such that $x_i \not\equiv 0, x_i \not\equiv x_{ij} \mod \mathfrak{m_i}$ for every $j$. By the Chinese Remainder Theorem, there is $x \in R^{\times}$, such that $x \equiv x_i \mod \mathfrak{m_i}$ for every $i$, which contradicts the above inclusion. As result, $R$ has property (U) and Claim 3 yields the conclusion.

The above corollary generalizes Lemma 2.9 of Apoorva Khare's preprint to semilocal rings.