Consider a composition of $ n$ into $ k$ parts with the length of blocks from left being $ [l_1,l_2,\dots,l_k]$. Then we select one from each block, which can be done in $ l_1\cdots l_k$ ways. The summation in the problem counts this for every composition. Equivalently, we can find the number of compositions with $ 2k-1$ selections where an odd numbered selector gives the selected element of a block and an even numbered selector gives the end of the next block (right after that selector). But selecting the last element of a block shall lead to a pair of odd and even selector to lie on the same element. Thus we obtain, $ \sum_{i=0}^{k-1}{k-1\choose i}{n\choose 2k-1-i}$

Your sum simplifies to $ \binom{n + k - 1}{2k - 1}$.