Alternatively, one can get explicit expression for a fixed $r$ via generating functions by noticing that the given sum equals
$$[y^{n+a}z^r]\ (y+1+z)^n (1+y(1+z))^n (1+z)^{-n}$$
and rewriting it as
$$[y^{n+a}]\ [z^r]\ \big((1+y)^2 - \frac{z}{1+z}y\big)^n.$$
Then for $r>0$ we evaluate it as
$$[y^{n+a}]\ \sum_{i=0}^{r} (-1)^i\binom{-i}{r-i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=1}^{r} \binom{r-1}{i-1} \binom{2(n-i)}{n+a-i}.$$