Let $M$ be a finitely generated torsion module over a PID $D$. Then by the structure theorem we have
$$M\cong \bigoplus_{i=1}^n D/(d_i),\quad\text{$d_i$ prime and $d_i\mid d_{i+1}$.}$$
Proposition. Any quotient of $M$ has the form $\bigoplus_i D_i$, where each $D_i$ is a submodule of $D/(d_i)$.
Proof. The bulk of the proof happens in the case where $M$ is primary, which I'll first deal with. Then the $d_i$'s are all powers $p^{e_i}$ of some fixed prime $p$, and $e_i\le e_{i+1}$. We can view $M$ as a tuple $(e_1,\ldots,e_n)$.
Let $N\cong M/M'$ be any quotient. Then $N$ is also a $p$-primary module, and admits a similar decomposition $N\cong \bigoplus_{j=1}^m D/(p^{f_j})$, where $f_j\le f_{j+1}$. We can also treat $N$ as a tuple $(f_1,\ldots,f_m)$. Now, the argument would be greatly expedited if $m=n$; so if for instance $n>m$, then I will instead consider the decomposition
$$N\cong \bigoplus_{j=1}^n D/(p^{f_j}),\quad\text{$f_j=0$ for $m<j\le n$.}$$
Likewise if $m>n$1, then I consider the decomposition
$$M\cong \bigoplus_{i=1}^m D/(p^{e_i}),\quad\text{$e_i=0$ for $n<i\le m$.}$$
Identifying $M$ and $N$ with their respective tuples, this corresponds to padding the shorter tuple with 0s to make them equal length. I would like to reiterate that identifying $p$-primary modules with tuples is possible because of the uniqueness of the decomposition (modulo the extra 0s).
From now I'll assume that $m=n$. Then what we want to prove is that
$$(f_1,\ldots,f_n)\le(e_1,\ldots,e_n),\quad\text{i.e. $f_i\le e_i$ for all $i$.}$$
For example, if $M=\Z/(p)\oplus\Z/(p^2)$, then the only nontrivial possibilities for $N$ are
$$\Z/(p)\oplus\Z/(p),\quad \Z/(p^2)\cong\Z/(1)\oplus\Z/(p^2),\quad \Z/(p)\cong\Z/(p)\oplus\Z/(1).$$
Let us proceed with the proof. The idea is that certain operations on $p$-primary modules can be interpreted in terms of tuples:
$p^kM$ is the module $\bigoplus_i D/(p^{e_i-k})$. Thus it is represented by the tuple $(e_1-k,\ldots,e_n-k)$2.
$p^kM/p^{k+1}M$ is the module
$$\bigoplus_i D/(p^{(e_i-k)-(e_i-k-1)}) = \bigoplus_i D/(p^{[e_i>k]}),$$
where the latter expression uses Iverson bracket notation. Note that it is always $\le(1,1,\ldots,1)$, and this implies that $p^kM/p^{k+1}M$ is a direct sum of $D/(p)$'s, and the number of nonzero summands is the dimension (as a vector space) over the field $D/(p)$.
We can go a step further. By identifying modules with tuples, we have the identity $p^kM = p^kM/p^{k+1}M + p^{k+1}M$. By induction we thus have the wonderful expression
$$M = M/pM + pM/p^2M + \ldots + p^{l-1}M/p^lM,\quad l=\max\,\{e_i\}.$$
Now let $l'=\max\,\\{f_j\\}$; there exists $x\in M$ with $\ann(x+M')=(p^{l'})$. Then
$$\ann(x+M')\supseteq \ann x\ge (p^l),$$
and thus $l'\le l$3. The rest of the proof boils down to the calculation
$$\begin{align*} N &= N/pN + \ldots + p^{l'-1}N/p^{l'}N \\ &\le M/pM + \ldots + p^{l'-1}M/p^{l'}M + \ldots + p^{l-1}M/p^lM \\ &= M. \end{align*}$$
We just need to show the
Lemma. $p^kN/p^{k+1}N$ is a quotient of $p^kM/p^{k+1}M$. Thus
$$\dim_{D/(p)} p^kN/p^{k+1}N \le \dim_{D/(p)} p^kM/p^{k+1}M,$$
and so
$$p^kN/p^{k+1}N \le p^kM/p^{k+1}M\ \ \text{as tuples.}$$
Proof. Suppose $N\cong M/M'$. We have the isomorphism $p^kN\cong p^kM/(p^kM\cap M')$ by applying the 1st iso theorem to the map
$$\begin{matrix} p^kM & \to & p^kN \\ m & \mapsto & \overline{m}. \end{matrix}$$
The rest is just spamming the isomorphism theorems:
$$\begin{align*} p^kN/p^{k+1}N &\cong {p^kM/(p^kM\cap M') \over p^{k+1}M/(p^{k+1}M\cap M')} \\ &\cong {p^kM/(p^kM\cap M') \over [p^{k+1}M+(p^kM\cap M')]/(p^kM\cap M')} \quad\text{(2nd iso theorem on $p^{k+1}M$ and $p^kM\cap M'$)}\\ &\cong {p^kM \over [p^{k+1}M+(p^kM\cap M')]} \\ &\cong {p^kM/p^{k+1}M \over [p^{k+1}M+(p^kM\cap M')]/p^{k+1}M}.\quad\qed \\ \end{align*}$$
This completes the proof for primary modules. In the general case, consider the decomposition of $M$ into primary components, $M\cong \bigoplus_i M_{p_i}$. Let $l_1,\ldots,l_n$ be the maximum multiplicities of each prime. For each $M'$, we have that
$${p_2}^{l_2}\cdots{p_n}^{l_n}(M/M') \cong (M/M')_{p_1}$$
but this is also isomorphic to the quotient
$${p_2}^{l_2}\cdots{p_n}^{l_n}M / ({p_2}^{l_2}\cdots{p_n}^{l_n}M\cap M') \cong M_{p_1}/(M_{p_1}\cap M').$$
Thus we can apply the previous part of the proof to conclude $(M/M')\_{p_1}\le M\_{p\_1}$ as tuples. Similarly, $(M/M')\_{p\_k}\le M\_{p\_k}$ for each $k$, and a moment's reflection should convince the reader that this implies the original proposition. $\qed$
1: It turns out that $m\le n$ always, but we don't know that at this juncture. 2: Here $a-b$ is defined as 0 if $a<b$, and the usual difference otherwise. 3: This is obvious in the case $D=\Z$: here $l$ denotes the maximum order of an element, and clearly passage to the quotient cannot increase the order.