It equals $$ \binom{2m}{m+k}\frac{3m-1-2k^2}{2m-1}=(m-1)\binom{2m}{m+k}+4m\binom{2m-2}{m+k-1}. $$$$ \binom{2m}{m+k}\frac{3m-1-2k^2}{2m-1}=-(m-1)\binom{2m}{m+k}+4m\binom{2m-2}{m+k-1}. $$ I got it by expanding $3m-1-2k^2=2(m^2-k^2)-(2m^2-3m+1)=2(m-k)(m+k)-(2m-1)(m-1)$.
