Because for each committee of size $k$ there are $\binom {k}{m}$ committees of size $m$. So you need to add $\binom {k}{m}$ as many as the number of committees with size $k$. You have:
$$ \binom {k}{m} + \binom {k}{m} + \dots + \binom {k}{m} $$ This addition has $\binom {n}{k}$ terms, it sums up to $\binom {n}{k}\binom {k}{m}$