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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 $$[y^{n+a}z^r]\ \big((1+y)^2 + \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}{i} (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i.$$ Then for $r>0$ we evaluate it as $$[y^{n+a}]\ \sum_{i=0}^{r} \binom{n}i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=1}^{\lfloor r/2\rfloor} \binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$

Alternatively, one can get an 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 $$[y^{n+a}z^r]\ \big((1+y)^2 + \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}{i} (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i,$$ which we evaluate as $$[y^{n+a}]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$

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 $$[y^{n+a}z^r]\ \big((1+y)^2 - \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^r \binom{n}{i} (-1)^i (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i.$$ Then for $r>0$ we evaluate it as $$[y^{n+a}]\ \sum_{i=0}^{r} \binom{n}i (-1)^i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = \sum_{i=1}^{\lfloor r/2\rfloor} (-1)^{r+i}\binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$

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 $$[y^{n+a}z^r]\ \big((1+y)^2 + \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}{i} (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i.$$ Then for $r>0$ we evaluate it as $$[y^{n+a}]\ \sum_{i=0}^{r} \binom{n}i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=1}^{\lfloor r/2\rfloor} \binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$

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 $$[y^{n+a}z^r]\ \big((1+y)^2 - \frac{z}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^r \binom{n}{i} (-1)^i (1+y)^{2(n-i)} \left(\frac{z}{1+z}y\right)^i.$$ Then for $r>0$ we evaluate it as $$[y^{n+a}]\ \sum_{i=0}^{r} \binom{n}i (-1)^i\binom{-i}{r-i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=1}^{r} \binom{n}i \binom{r-1}{i-1} \binom{2(n-i)}{n+a-i}.$$

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 $$[y^{n+a}z^r]\ \big((1+y)^2 - \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^r \binom{n}{i} (-1)^i (1+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i.$$ Then for $r>0$ we evaluate it as $$[y^{n+a}]\ \sum_{i=0}^{r} \binom{n}i (-1)^i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = \sum_{i=1}^{\lfloor r/2\rfloor} (-1)^{r+i}\binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$