Let $V$ be a finite-dimensional representation of a finite group $G$. A composition series of $V$ is a filtration $0 = V_0 \subseteq V_1 \subseteq \cdots \subseteq V_n = V$ of subspaces such that each $V_i$ is a submodule, and each quotient $V_{i} / V_{i-1}$ (called a composition factor) is a simple module. The Jordan-Hölder theorem for representations applies, which states that although there may be many composition series for $V$, the list of composition factors and their multiplicities is always the same. Therefore we get a useful invariant of $V$: which simples appear in a composition series, with what multiplicity. This might be what people mean when they say that simple modules are the "building blocks" of modules.
In general, this list does not classify $V$ up to isomorphism: there could be many non-isomorphic modules with the same composition factors. But if $V$ is semisimple (for example, when the characteristic of the field does not divide $|G|$), then that list does classify $V$ up to isomorphism. (It is easy to check that if $V \cong S_1 \oplus \cdots \oplus S_n$ where each $S_i$ is simple, then any composition series of $V$ has composition factors given by the $S_i$ in some order).
For an example where a module is not semisimple, consider the group $G = \{1, g\}$ of order $2$ acting on the vector space $V = \mathbb{F}_2\{e_1, e_2\}$ of $2$-element vectors over the finite field $\mathbb{F}_2$, where $g$ acts by switching $e_1$ and $e_2$. We can write out all of the $G$-submodules explicitly, just by checking the orbits of the four vectors $0, e_1, e_2, e_1 + e_2$:
$$ \{0\}, \quad V_1 = \{0, e_1 + e_2\}, \quad V,$$
so there is a unique 1-dimensional submodule, isomorphic to the trivial module, which I've called $V_1$. Therefore a composition series for $V$ is $0 \subseteq V_1 \subseteq V$, with composition factors $V_1$ and $V / V_1$ both isomorphic to trivial modules. However $V$ is not isomorphic to a direct sum of two trivial modules since there is a unique $1$-dimensional submodule of $V$. If we had another two-dimensional module $U$ which had trivial modules as its composition factors, we would still need more information to tell whether it were isomorphic to $V$ or not.
The condition is equivalent to the property that there is some cardinal $\kappa$ such that every left $R$-module is a direct sum of modules with at most $\kappa$ generators.
Theorem 2 of
Warfield, R. B. jun, Rings whose modules have nice decompositions, Math. Z. 125, 187-192 (1972). ZBL0218.13012.
shows that the commutative rings with this property are precisely the Artinian principal ideal rings.
For not necessarily commutative rings, Theorem 2.2 of
Griffith, P., On the decomposition of modules and generalized left uniserial rings, Math. Ann. 184, 300-308 (1970). ZBL0175.31703.
shows that all rings with this property are left Artinian.
There is a property of rings called (left) pure semisimplicity, which has several equivalent definitions. Section 4.5 of
Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.
gives a good survey, including such results as:
(Theorem 4.5.4) A left pure semisimple ring is left Artinian, and every left $R$-module is a direct sum of indecomposable finite length indecomposable modules.
(Theorem 4.5.7) $R$ is left pure semisimple if and only if every left $R$-module is a direct sum of indecomposable modules, if and only if there is a cardinal $\kappa$ such that every left $R$-module is a direct sum of modules of cardinality less than $\kappa$.
(Note that one consequence of all of this is that if every left $R$-module is a direct sum of modules with at most $\kappa$ generators for some $\kappa$, then in fact every left $R$-module is a direct sum of finitely generated modules.)
So a complete answer to the question is that the rings in question are precisely the left pure semisimple rings, although one might argue that this is just replacing the original condition with an equally mysterious one.
But there is a conjecture (the Pure Semisimplicity Conjecture, which is 4.5.26 in Prest's book) that the left pure semisimple rings are precisely the rings of finite representation type (i.e., for which every module is a direct sum of indecomposable modules, and with only finitely many indecomposables). It is known that every ring of finite representation type is left pure semisimple. Also, "finite representation type" is a left/right symmetric property, so if the conjecture is true then pure semisimplicity is a left/right symmetric property.
Best Answer
Here's an elementary linear algebra proof.
Let $M$ be a $k[x]/(x^2)$-module, and let $T:M\to M$ be multiplication by $x$.
Let $U=T(M)$. Since $x^2=0$, $T(M)\leq\ker(T)$, and so we can choose a subspace $V$ of $M$ such that $\ker(T)=U\oplus V$.
Then we can choose $W$ so that $M=U\oplus V\oplus W$.
Then $T$ induces an isomorphism from $W$ to $U$. Let $\{w_i\}$ be a basis of $W$, so $\{w_ix\}$ is a basis of $U$. Let $\{v_j\}$ be a basis of $V$.
Then $M$ is the direct sum of submodules $V_j=\langle v_j\rangle$, which are isomorphic to $k[x]/(x)$, and $L_i=\langle w_i,w_ix\rangle$, which are isomorphic to $k[x]/(x^2)$.