Here is a proof repeating the trick I used in this <a href="http://mathoverflow.net/questions/143334/proving-sum-k-02m-1k-binom2mk3-1m-binom2mm-binom3mm/144067#144067">earlier answer</a>. By negating the second binomial coefficient, the identity in the question is equivalent to

$$ \sum_{k=0}^n (-1)^k \binom{4n+1}{k} \binom{-(2n+1)}{n-k} = 2^{2n} \binom{2n}{n}. $$

This is the special case $x= 4n+1$ of the following identity

$$ \sum_{k=0}^n (-1)^k \binom{x}{k}\binom{2n-x}{n-k} = (-1)^n 2^{2n} \binom{(x-1)/2}{n}. $$

When $x=2r+1$ is odd the left-hand side is zero since the summands for $k=r-j$ and $k=r+1+j$ cancel. So the two sides agree when $x=1,3,\ldots,2n-1$. When $x = 2n+1$ the left-hand side is $\sum_{k=0}^n (-1)^n \binom{2n+1}{k} = (-1)^n 2^{2n+1}/2 = (-1)^n 2^{2n}\binom{n}{n}$. So the two sides agree at $n+1$ points. Since they are polynomials of degree $n$ in $x$, they must agree for all $x$.