The Statement
I suspect the following proposition is well known, but I found no reference.
Proposition If $A$ is a principal ideal domain, if $I$ is a nonzero ideal of $A$, and if $M$ is an $A/I$-module, then $M$ is a direct sum of finitely generated submodules.
The Proof
Proof (sketch) By the Chinese Remainder Theorem we can assume that $I$ is of the form $(p^n)$ where $p$ is a prime element of $A$ and $n$ a positive integer. Let $K$ be the residue field $A/(p)$, and let $S\subset A$ be system of representatives of the classes mod $(p)$ such that $0\in S$. We shall prove:
(a) There is a subset $\widetilde B=\bigsqcup_{i=0}^{n-1}\widetilde B_i$ of $M$ such that each element of $M$ can be written in a unique way as $$ \sum\ s(i,\widetilde b_j)\ p^i\ \widetilde b_j $$ where the sum runs over those pairs $(i,\widetilde b_j)$ with $0\le i\le j < n$, $\widetilde b_j\in B_j$, the $(s(i,\widetilde b_j))$ forming a finitely supported family of elements of $S$.
Clearly (a) implies the proposition. We sketch a proof of (a).
The multiplication by $p$ induces $K$-linear surjections $$ M/pM\xrightarrow{f_1}pM/p^2M\xrightarrow{f_2}\cdots\xrightarrow{f_{n-2}}p^{n-2}M/p^{n-1}M\xrightarrow{f_{n-1}}p^{n-1}M. $$ Pick a $K$-basis $B_0$ of $\operatorname{Ker} f_1$,
$\bullet$ complete $B_0$ to a $K$-basis $B_0\sqcup B_1$ of $\operatorname{Ker}(f_2\circ f_1)$,
$\bullet$ complete $B_0\sqcup B_1$ to a $K$-basis $B_0\sqcup B_1\sqcup B_2$ of $\operatorname{Ker}(f_3\circ f_2\circ f_1)$,
$\bullet\ \cdots$,
$\bullet$ complete $B_0\sqcup\cdots\sqcup B_{n-3}$ to a $K$-basis $B_0\sqcup\cdots\sqcup B_{n-2}$ of $\operatorname{Ker}(f_{n-1}\circ\cdots\circ f_1)$, and
$\bullet$ complete $B_0\sqcup\cdots\sqcup B_{n-2}$ to a $K$-basis $B_0\sqcup\cdots\sqcup B_{n-1}$ of $M/pM$.
Let $\widetilde b_j\in M$ be a lift of $b_j\in B_j\subset M/pM$, and let $\widetilde B_j\subset M$ be the set of all $\widetilde b_j$.
We claim that
(b) the $\widetilde B_j$ satisfy (a).
Clearly
(c) the map $f_{k-1}\circ\cdots\circ f_1$ induces a $K$-linear isomorphism $$ \frac{M/pM}{\operatorname{Ker}(f_{k-1}\circ\cdots\circ f_1)}\simeq\frac{p^kM}{p^{k+1}M}\quad. $$ Let $\sigma_k:p^kM\to p^kM/p^{k+1}M$ be the canonical projection. It is easy to see that (c) implies
(d) $\sigma_k\left(p^k\left(\widetilde B_k\sqcup\cdots\sqcup\widetilde B_{n-1}\right)\right)$ is a $K$-basis of $p^kM/p^{k+1}M$.
Taking (d) for granted, we prove the uniqueness of the family $(s(i,b_j))$. (As often a close look at the proof of uniqueness suggests a proof of existence.)
Assume by contradiction we have distinct families $(s(i,b_j))$ and $(t(i,b_j))$ as above such that $$ \sum\ s(i,b_j)\ p^i\ \widetilde b_j=\sum\ t(i,b_j)\ p^i\ \widetilde b_j. $$ Let $k$ be the least nonnegative integer for which there is a $b_j$ with $s(k,b_j)\ne t(k,b_j)$. The congruence $$ \sum\ s(k,b_j)\ p^k\ \widetilde b_j\equiv\sum\ t(k,b_j)\ p^k\ \widetilde b_j\bmod p^{k+1}, $$ where the sums run over $k\le j < n$ and $b_j\in B_j$, contradicts (d). QED
The Question
My question has five parts:
(e) Is the above proposition true?
(f) Is it well known?
(g) Do you know any reference?
(h) Is the above argument correct?
(i) Is there a better argument?
Yes, this is very old work. In 1935, Koethe proved that the modules of Artinian principal ideal rings are all direct sums of cyclic submodules. Of course, your example falls into this category. In fact, all of the proper quotients of a principal ideal domain are Artinian principal ideal rings (in fact they are also self-injective, hence quasi-Frobenius.)
If you enjoy reading in German, you can look up
Of course since then, there have been many proofs written in English :) My go-to references for serial rings are anything by Puninski, such as this article or even his textbook Serial rings.
Eisenbud and Griffith's article is also freely available: Serial rings
Rings for which all right modules are decomposable into direct sums of finitely generated modules are called right pure-semisimple rings. They are pretty well-studied, and the place I would begin studying them is Algebra, K-theory, Groups, and Education: On the Occasion of Hyman Bass's 65th birthday