For fixed $r$ you may expand ${2k-n+a\choose r}$ as a polynomial in $k$ (whose coefficients are polynomials in $n-a$) as ${2k-n+a\choose r}=\sum_{i=0}^r c_i {k\choose i}$. To find $c_i$'s, fix $s\in \{0,1,\ldots,r\}$; substitute $k=j\in \{0,1,\ldots,s\}$ to get $\sum_{i=0}^j c_i{j\choose i}={2j-n+a\choose r}$, multiply this by $(-1)^{s-j}{s\choose j}$ and sum up over $j=0,\ldots,s$. Then $c_i$ comes with the coefficient $$\sum_{j=i}^s (-1)^{s-j}{j\choose i}{s\choose j}=\sum_{j=i}^s (-1)^{s-j}{s\choose i}{s-i\choose j-i}={s\choose i}\sum_{\ell=0}^{s-i}(-1)^{s-i-\ell}{s-i\choose \ell}=\delta_{is},$$ and we get $$c_s=\sum_{j=0}^s (-1)^{s-j}{s\choose j}{2j-n+a\choose r}.$$
Therefore your sum $S$ equals $$S=\sum_{i=0}^r c_i\sum_{k=0}^n{n\choose k}{k\choose i}{n\choose k+a}.$$ Now use $${n\choose k}{k\choose i} ={n\choose i} {n-i\choose k-i},$$ and finally apply Vandermonde convolution $$\sum_{k=i}^n {n-i\choose k-i}{n\choose k+a}={2n-i\choose n+a}$$ to express your sum as $$S=\sum_{i=0}^rc_i{n\choose i}{2n-i\choose n+a}.$$