Vandermonde's identity gives $$\sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.$$
Here is an example of Vandermonde's-like identity: For all $0 \le m \le n$,
$$\sum_{k=0}^{2m} \binom{\left\lfloor\frac{n+k}{2}\right\rfloor}{k}\binom{m+\left\lfloor\frac{n-k}{2}\right\rfloor}{2m-k}=\binom{m+n}{2m}$$
(Note that $\left\lfloor\frac{n+k}{2}\right\rfloor+\left(m+\left\lfloor\frac{n-k}{2}\right\rfloor\right)$ is either $m+n$ or $m+n \pm 1$)
I wonder if there are some similar identities where $m(k)$ and $n(k)$ are functions of $k$ and $m(k)+n(k)$ is 'almost' constant, says $m+n$, the identity looks like
$$\sum_{k=0}^r \binom{m(k)}{k}\binom{n(k)}{r-k}=\binom{m+n}{r}?$$
Best Answer
Here are some almost "Vandermonde-like" identities that may be of interest. They're not exactly what you're asking for, but they are pretty close.
$$\begin{align*} \sum_{k=0}^n \binom{p+k}{k} \binom{q+n-k}{n-k} &= \binom{n+p+q+1}{n} \\ 2 \sum_{k=0}^r \binom{n}{2k} \binom{n}{2r+1-2k} &= \binom{2n}{2r+1} \\ 2 \sum_{k=0}^r \binom{n}{2k} \binom{n}{2r-2k} &= \binom{2n}{2r} + (-1)^k \binom{n}{r} \\ 2 \sum_{k=0}^{r-1} \binom{n}{2k+1} \binom{n}{2r-2k-1} &= \binom{2n}{2r} - (-1)^k \binom{n}{r} \end{align*}$$
The first one is on p. 148 of Riordan's Combinatorial Identities, and the last three are on p. 144. There may be more in Riordan's book; I just flipped through until I found a few.