I expect that this question is an elementary exercise in combinatorics, so hopefully somebody who knows more than me can explain.

Specifically, if $m\in\mathbb{N}$ and $$f(x)=\sum_{n=1}^{\infty}a_nx^n,$$ one obviously has $$f^m(x)=\sum_{n=m}^{\infty}\left(\sum_{k_1+k_2+\cdots + k_m=n}a_{k_1}a_{k_2}\cdots a_{k_m}\right)x^n.$$

The number of distinct products appearing in the iterated convolution is the integer partition number $p(n-m)$ so, writing the inner sum as

(1) $$\sum_{j_1+2j_2+\cdots + mj_m=n}s_{n}(j_1,j_2,...,j_m)a^{j_1}_1a^{j_2}_2\cdots a^{j_m}_m,$$ the questions arise:

What is the coefficient sequence of integers $s_{n}(j_1,j_2,...,j_m)$, explicitly?

What is its combinatorial significance?

EDIT: The fact that the number of terms in the sum is $p(n-m)$ is perhaps not so obvious, but you can reason as follows: note that the largest subscript that can appear in the coefficient of $x^n$ is $n-m+1$, because $n-m+1+(m-1)\times 1=n$. This restricts you to having only those partitions of $n$ whose largest part is $m-n+1$, that is, the set of solutions to the diophantine equation

(2)$$ j_1+2j_2+\cdots +(n-m+1)j_{n-m+1}=n.$$ On the other hand, since there are a total of $m$ coefficients in each product (counting repetitions), you know that

(3)$$j_1+j_2+\cdots +j_{n-m+1}=m.$$ Thus, subtracting (2) from (1), you get $$j_2+2j_3+\cdots +(n-m)j_{n-m+1}=n-m,$$ the number of distinct solutions to which is the partition number $p(n-m)$. I am trying to think of a better notation for the summation in (1), as I think it is definitely misleading.

EDIT: I should've indexed everything by $+1$ so that (1) becomes $$\sum_{j_1+2j_2+\cdots +(n-m)j_{n-m}=n-m}s(j_1,j_2,...,j_{n-m})a_1^{m-(j_1+\cdots +j_{n-m})}a_2^{j_1}\cdots a_{n-m+1}^{j_{n-m}}.$$