I just saw this question. If I'm not missing something, I think I have a quick answer to your question asking for a combinatorial interpretation for $a_{n,k}$.

Edit: I just realized that my answer is closely related to Gjergji's answer -- I hadn't read his closely enough before. But what I would add to his interpretation is that one may think of his recursive process as giving chains of a particular length in the partition lattice (as detailed below). So you might tap into the literature on the partition lattice.

This number seems to count the chains $\hat{0} = u_k < u_{k-1} < \cdots < u_1 < \hat{1}$ in the partition lattice $\Pi_n $, namely the poset of set partitions of {1,...,n} ordered by reverse refinement (i.e. with $u \le v$ iff $v$ is obtained from $u$ by merging blocks). I've indexed in reverse to align with your notation. Notice that going up a cover relation reduces the number of blocks by exactly one, so $S(n,n-j)$ counts poset elements of rank $j$. Once you choose $u_{k-1}$ of rank $j$, you might as well think of each block in $u_{k-1}$ as a letter, so that counting ways to go up from your chosen $u_{k-1}$ which has rank $j$ to any element $u_{k-2}$ of rank $j'$ satisfying $u_{k-1} < u_{k-2}$, you have $S(n-j,n-j')$ choices for this $u_{k-2}$, etc. Thus, your summands $1 = m_1 < \cdots < m_k = n$ are a choice of which ranks in the partition lattice to use for your chain, and then your products are progressively calculating the ways to choose the next element in a chain having the appropriate rank and above the lower elements of the chain, given the choices so far of lower elements in the chain. In particular, $S(m_j,m_{j-1})$ seems to count the choices for $u_{j-1}$ of rank $m_{j-1}$ once you have chosen the chain elements $u_k,\dots ,u_j$ that are all below it in the chain.

Hopefully I didn't mess up any indexing, but I easily could have. There is a large literature regarding the partition lattice, so maybe it's worth seeing what's been done with chain enumeration there.

I just saw this question. If I'm not missing something, I think I have a quick answer to your question asking for a combinatorial interpretation for $a_{n,k}$.

This number seems to count the chains $\hat{0} = u_k < u_{k-1} < \cdots < u_1 < \hat{1}$ in the partition lattice $\Pi_n $, namely the poset of set partitions of {1,...,n} ordered by reverse refinement (i.e. with $u \le v$ iff $v$ is obtained from $u$ by merging blocks). I've indexed in reverse to align with your notation. Notice that going up a cover relation reduces the number of blocks by exactly one, so $S(n,n-j)$ counts poset elements of rank $j$. Once you choose $u_{k-1}$ of rank $j$, you might as well think of each block in $u_{k-1}$ as a letter, so that counting ways to go up from your chosen $u_{k-1}$ which has rank $j$ to any element $u_{k-2}$ of rank $j'$ satisfying $u_{k-1} < u_{k-2}$, you have $S(n-j,n-j')$ choices for this $u_{k-2}$, etc. Thus, your summands $1 = m_1 < \cdots < m_k = n$ are a choice of which ranks in the partition lattice to use for your chain, and then your products are progressively calculating the ways to choose the next element in a chain having the appropriate rank and above the lower elements of the chain, given the choices so far of lower elements in the chain. In particular, $S(m_j,m_{j-1})$ seems to count the choices for $u_{j-1}$ of rank $m_{j-1}$ once you have chosen the chain elements $u_k,\dots ,u_j$ that are all below it in the chain.

Hopefully I didn't mess up any indexing, but I easily could have. There is a large literature regarding the partition lattice, so maybe it's worth seeing what's been done with chain enumeration there.