Brian M. Scott's answer is a nice one for a rearrangement of the identity; here's a perhaps less satisfying argument for the identity as written.
Assume $2k\leq n$ so that $n-2k$ is nonnegative. Let $S = \{1, \ldots, n\}$, and let ${S \choose k}$ denote the set of size-$k$ subsets of $S$, so that $\left\lvert {S \choose k} \right\rvert = {n \choose k}$. Let's suppose (here's the less satisfying part) that we have in hand some bijection $f : {S \choose k} \to {S \choose k}$ such that $t \cap f(t) = \emptyset$ for all $t \in {S \choose k}$. When $n = 2k$ we could just take $f(t) = S - t$, and for general $k \leq n/2$ it can be shown that such a bijection exists, but it seems surprisingly hard to actually construct a natural example of such a bijection when $k < n/2$. (Is there an obvious example I'm missing?)
Assuming we have the bijection, here's an argument for the identity. From our set of $n$ people, we wish to choose a team $t$ of $k$ people, as well as a supervisor that is not in $t \cup f(t)$. (In particular, and in contrast to the "captains" in Brian M. Scott's answer, we view the supervisor as not being a member of the team.)
If we choose the team $t$ first, then there are ${n \choose k}$ ways to choose the team, and, since $t \cap f(t) = \emptyset$, there are exactly $2k$ choices of supervisor that are excluded, so there are $(n - 2k){n \choose k}$ ways to choose both the team and the supervisor.
If we choose the supervisor first, say $v$ is the supervisor, then we have $n$ possible choices for $v$. There are ${n-1 \choose k}$ ways to pick a team $t$ not containing the supervisor, but we also need to enforce the constraint that $v \notin f(t)$. Any team $t$ violating this constraint must have $f(t) = \{v\} \cup q$, where $q$ is a set of $k-1$ people other than $v$. With ${n-1 \choose k-1}$ choices for $q$, and with each choice of $q$ giving exactly one forbidden $t$ since $f$ is a bijection, this gives ${n-1 \choose k} - {n-1 \choose k-1}$ valid choices of team for the given supervisor, so gives $n\left[{n-1 \choose k} - {n-1 \choose k-1}\right]$ ways to assemble the team and supervisor.