Let $S$ be a set of $n+r-1$ elements. Both sides count the number of ways to select two disjoint sets $A,B\subseteq S$ of size $r$ and possibly an element $c\in S \setminus B$.
We first observe that $(n-r)\binom{n}{r}=n\binom{n-1}{r}$ as both sides count the number of ways to form a team of size $r+1$ with a captain out of $n$ people.
Applying the above on the original LHS we get $n\binom{n-1}{r}\binom{n+r-1}{r}$, which corresponds to selecting $A$ ($r$ out of $n+r-1$), then $B$ ($r$ out of the remaining $n-1$) and $c$ ($n-1$ choices in $S\setminus B$ and one option of not choosing $c$).
The RHS argument goes as follows: choose $2r$ elements for both $A$ and $B$, then choose $r$ of them to make $B$. Then, as before, there are $n$ options of choosing $c\in S\setminus B$ or none at all.