I guess that $r \le n$. Then it seems to me that you could proceed as follows: regarding integers as sets (and binomial coefficients as sets of sets, and sums of integers as disjoint unions of sets) in a hopefully obvious way, map $\sum_{R = 0}^N \binom R r\binom{N - R}{n - r}$ to $\binom{N + 1}{n + 1}$ by sending the element $A \times B$ of the $R$th summand to $A \cup \{R + 1\} \cup \{b + R + 2 : b \in B\}$. The inverse map $\binom{N + 1}{n + 1} \to \sum_{R = 0}^N \binom R r\binom{N - R}{n - r}$ sends $\{i_1, \dots, i_r, i_{r + 1}, i_{r + 2}, \dotsc, i_n\}$ to $\{i_1, \dotsc, i_r\} \times \{i_{r + 2}, \dotsc, i_n\}$, viewed as an element of $\binom{\{1, \dotsc, i_{r + 2} - 1\}}r \times \binom{\{i_{r + 2} + 1, \dotsc, N\}}{n - r}$.
EDIT: After following @MartinSleziak's link, I think that this is just Brian M. Scott's answer. I leave it here in case it is an acceptable answer to the question, but mark it community wiki to avoid reputation.