$$ n\ge 4, S(n,n-2)=\binom{n}{3}+\frac{1}{2}\binom{n}{2}\binom{n-2}{2} $$
Proving this combinatorially. I understand what the LHS is doing. There are a couple of things I don't follow on the right-hand side and why they are relevant. Why are we taking half of that product, what is the relevance on $n$ choose $3$?
The Stirling number $S(n,n-2)$ is the number of ways of partitioning a set of $n$ elements into $n-2$ non-empty subsets.
Because $n-2$ and $n$ are quite close, most of the sets have to be singletons. Indeed, there are only two cases: either all sets are size $1$, except for one set of size $3$ (this can be done in $\binom n3$ ways), or there are two sets of size $2$. These two sets of size $2$ can be formed in $\tfrac12\binom n2\binom {n-2}2$ ways.
Thus, your formula follows.