Prove that $\sum_{k=0}^m \binom{n}{k} \binom{n-k}{m-k}= 2^m \binom{n}{m}$ for $m < n $

You have $$ \binom{n}{k} \binom{n-k}{m-k} = \frac{n!}{k!(m-k)!(n-m)!} = \binom{n}{m} \binom{m}{k}, $$ so $$ \sum_{k=0}^m \binom{n}{k} \binom{n-k}{m-k} = \binom{n}{m} \sum_{k=0}^m \binom{m}{k} = 2^m \binom{n}{m}. $$

PS: You can also interpret this formula combinatoricaly: The right side $2^m \binom{n}{m}$ tells us that we have $n$ colourless balls, out of which we choose $m$ many (giving us $\binom{n}{m}$), and paint each of them either black or white (giving us $2^m$). The term $\binom{n}{k} \binom{n-k}{m-k}$ on the left side tells us that we choose $k$ balls to paint black and then $m-k$ balls to paint white; summing this over $k = 0, 1, \dotsc, m$ gives us the same result as before.

Another solution is by double counting.

Suppose that we have n balls, and we want to choose m balls among them and color each of them by blue or red. In how many ways we can do that?

Right side: We can simply choose m of n balls ( $n \choose m$ ) and then choose a color for each of these m balls ($2^m$).

Left side: We can choose $k$ balls to be colored by blue ( $n \choose k$ ) and the m-k balls from the remaining balls to be colored by red ( $n-k \choose m-k$ ) and we must do so for all $1\leq k \leq m$.