One way to interpret the numbers in Pascal’s triangle is that $n\choose k$ is the number of different paths from the point $(0,0)$ at the top of the triangle (the $0$-th number in row #$0$, where there is a $1$ that equals $0\choose0$) to the point $(n,k)$ (where $n\choose k$ appears) using steps that either increase $n$ by $1$ and leave $k$ alone or that increase both $n$ and $k$ by $1$.
In this picture, the possible steps in the path would move by one square to the east (increasing only $n$) or to the southeast (increasing both $k$ and $n$).
If you’re patient and careful, you can check, for example, that there are $20$ different paths from the green square to the yellow square by southward or southeastward steps.
It’s not hard to see that every $S$ or $SE$-moving path from the green square to the yellow square has to pass through one of the blue squares, where the $n$ value is smaller than in the yellow square by $2$. From either the top or bottom blue square, there’s only one way to continue on to the yellow square, but from the middle blue square, there are two ($S$, then $SE$ or vice versa).
This is pretty much a proof of your identity, because the number of paths from the green square to any (yellow) square containing $n\choose k$ must pass through one of exactly three (blue) squares in the column where $n$ is smaller by $2$, and the paths to the yellow square can be counted separately according to which blue square they pass through, and if you do so, you get the formula you asked about.