A previous post characterised the quotient modules. It turns out that submodules admit the exact same characterisation---the proof is different but it still shares some elements. I'll refer to the other post as QUOT.
Proposition. Any submodule of $M\cong \bigoplus_i D/(d_i)$ has the form $\bigoplus_i D_i$, where each $D_i$ is a submodule of $D/(d_i)$.
Proof. I handle the primary case first. Then
$$M\cong \bigoplus_{i=1}^n D/(p^{e_i}).$$
If $N$ is a submodule then it is also $p$-primary and
$$N\cong \bigoplus_{j=1}^m D/(p^{f_j}).$$
Using the padding trick in QUOT, we can assume that $m=n$. Let
$$l=\max\,\{e_i\},\quad l'=\max\,\{f_j\}.$$
I showed in QUOT that $l'\le l$, and thus $p^{l-l'}M$ is well-defined. The annihilator of $N$ is $(p^{l'})$, and $p^{l-l'}M$ is the largest submodule of $M$ whose annihilator is $(p^{l'})$.1 Thus $N$ is a submodule of $p^{l-l'}M$, whose largest invariant factor is $p^{l'}$. In other words, the maximum components of the tuples corresponding to $N$ and $p^{l-l'}M$ are both $l'$.
If $N>p^{l-l'}M$ as tuples, then there is some $k$ such that the tuple $p^kN$ has more nonzero components than $p^{l-l'+k}M$. For example, if
$$N = (4,3,3,1),\quad p^{l-l'}M = (4,3,2,1),$$
then
$$p^2N = (2,1,1,0),\quad p^{l-l'+2}M = (2,1,0,0).$$
Let $r$ and $s$ be the number of nonzero components in $p^kN$ and $p^{l-l'+k}M$ respectively. Then $p^kN$ contains the vector space $\bigl(D/(p)\bigr)^r$ and $p^{l-l'+k}M$ contains the vector space $\bigl(D/(p)\bigr)^s$; by the same argument concluding that $N\le p^{l-l'}M$, we conclude that
$$\bigl(D/(p)\bigr)^r \quad\text{embeds into}\quad \bigl(D/(p)\bigr)^s.$$
But this is impossible since $r>s$, and thus $N\le p^{l-l'}M\le M$, proving the primary case.
In the general case, consider the decompositions
$$M = \bigoplus_i M_{p_i},\quad N = \bigoplus_i N_{p_i}.$$
Since $M_{p_i}$ is the largest submodule of $M$ whose annihilator is contained in $(p_i)$, we have $N_{p_i}\subseteq M_{p_i}$ as submodules. Thus $N_{p_i}\le M_{p_i}$ as tuples, completing the proof. $\qed$