I'll abbreviate $\mathbb{Z}/p^k \mathbb{Z}$ to $L_k$. Let $G = \bigoplus_{i=1}^M L_i^{a_i}$. I claim that $\bigoplus_{i=1}^N L_i^{b_i}$ is in the quotient series of $G$ if and only if the Littlewood-Richardson coefficient
$$c_{(1^a_1),\ \ (2^{a_2}),\ \ \cdots,\ \ (M^{a_M})}^{(1^{b_1} 2^{b_2} \cdots N^{b_N})}$$
is nonzero. Note the parentheses and commas: We have distinct $M$ partitions on the bottom and only one partition on the top.

You can decide whether or not this is an explicit enough answer to be useful to you.

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<b>Proof</b> In general, let $\alpha$, $\beta$ and $\gamma$ be three partitions. Then there is a short exact sequence
$$0 \to \bigoplus L_{\alpha_i} \to \bigoplus L_{\beta_i} \to \bigoplus L_{\gamma_i} \to 0$$
if and only if $c_{\alpha \gamma}^{\beta}$ is nonzero. This is the same as asking that there be a semistandard Young tableaux of shape $\beta/\alpha$ whose rectification is a given tableaux of shape $\gamma$. See, for example, [Fulton's survey][1]. The short exact sequences of abelian groups are in Section 2; the tableaux are briefly mentioned in Section 9 and can be found in more detail in many places, such as Stanley's <i>Enumerative Combinatorics II</I> or Fulton's <i>Young Tableaux</i>.

In your situation, you want to know when we have $c_{T \alpha,\ \gamma}^{(T+1) \alpha}$ nonzero for $T$ large. So we are looking for tableaux of shape $(T+1)\alpha/T \alpha$ whose rectification is a given tableaux of shape $\gamma$. But, once $T$ is large enough, $(T+1)\alpha/T \alpha$ is just a union of disconnected rectangles. Specifically, if $\alpha = 1^{a_1} 2^{a_2} \cdots M^{a_M}$, then the connected components of $(T+1)\alpha/T \alpha$ are rectangles of shape $k \times a_k$, for $1 \leq k \leq M$. Standard properties of Young tableaux then give the answer above.


  [1]: https://arxiv.org/abs/math/9908012