Lemma IV.6.11 of Hungerford's Algebra is
Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and each $n_i > 0$, then there is an $R$-module isomorphism
$$R/(r) \cong R/(p_1^{n_1}) \oplus \cdots \oplus R/(p_k^{n_k}).$$
Consequently every cyclic $R$-module of order $r$ is a direct sum of $k$ cyclic $R$-modules of orders $p_1^{n_1},\ldots,p_k^{n_k}$ respectively.
In his sketch of a proof for Lemma IV.6.11, he says
The last statement of the lemma is an immediate consequence of the fact that $R/(c)$ is a cyclic $R$-module of order $c$ for each $c \in R$ by Theorem 6.4.
Theorem IV.6.4 goes
Theorem 6.4. Let $A$ be a left module over an integral domain $R$ and for each $a \in A$ let $\mathcal{O}_a = \{r \in R \mid ra = 0\}$.
(i) $\mathcal{O}_a$ is an ideal of $R$ for each $a \in A$.
(ii) $A_t = \{a \in A \mid \mathcal{O}_a \ne 0\}$ is a submodule of $A$.
(iii) For each $a \in A$ there is an isomorphism of left modules $$R/\mathcal{O}_a \cong Ra = \{ra \mid r \in R\}.$$
Let $R$ be a principal ideal domain and $p \in R$ a prime.
(iv) If $p^i a = 0$ (equivalently $(p^i) \subset \mathcal{O}_a$), then $\mathcal{O}_a = (p^j)$ with $0 \le j \le i$.
(v) If $\mathcal{O_a} = (p^i)$, then $p^j a \ne 0$ for all $j$ such that $0 \le j < i$.
I don't see how Theorem IV.6.4 implies that $R/(c)$ is a cyclic $R$-module of order $c$ for each $c \in R$. Couldn't there be a principal ideal domain $R$ and $c \in R$ such that for every $R$-module $A$, there is no $a \in A$ having $\mathcal{O}_a = (c)$?
No, because $A:=R/(c)$ is cyclic (generated by $\bar{1}$) and ${\cal O}_{\bar{1}}=(c)$. Check for yourself! (A scalar element $r\in R$ annihilates $1+(c)$ in $R/(c)$ means that .... [write stuff down, profit]).