This problem derives from an expression of probability in random walk. I hope to
prove that \begin{equation*} \sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)} \end{equation*}
It seems quite right, I verified this equation from $n=1$ to $6$ without any idea to prove it.
This equality is found by @Sangchul Lee.
The RHS is the coefficient of $z^{2n}$ in the product between
$$ \sum_{l\geq 0}\binom{2l}{l}(xz)^{2l}=\frac{1}{\sqrt{1-4x^2 z^2}}\qquad\text{and}\qquad \sum_{l\geq 0}\binom{2l}{l}(yz)^{2l}=\frac{1}{\sqrt{1-4y^2 z^2}} $$
i.e. the coefficient of $z^{2n}$ in $\frac{1}{\sqrt{(1-4x^2 z^2)(1-4y^2 z^2)}}$.
In the LHS we may recognize a Legendre polynomial:
$$ P_n(t)=\frac{1}{2^n}\sum_{l=0}^{n}\binom{l}{n}^2(t+1)^l (t-1)^{n-l} $$
hence the given identity turns out to be a consequence of the following identity, about the generating function for Legendre polynomials:
$$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n\geq 0}t^n P_n(x) $$
which can be proved through Rodrigues' formula or Bonnet's recursion formula.
We seek to verify that
$$\sum_{l=0}^{n} {n\choose l}^2 (x+y)^{2l} (x-y)^{2n-2l} = \sum_{l=0}^{n} {2l\choose l} {2n-2l\choose n-l} x^{2l}y^{2n-2l}.$$
Now we see on the LHS that the powers of $x$ and $y$ always add up to $2n$ and the exponent on $x$ determines the one on $y.$ Extracting the coefficient on $[x^q][y^{2n-q}]$ we obtain
$$\sum_{l=0}^{n} {n\choose l}^2 \sum_{p=0}^q {2l\choose p} (-1)^{2n-2l-(q-p)} {2n-2l\choose q-p} \\ = \sum_{l=0}^{n} {n\choose l}^2 \sum_{p=0}^q {2l\choose p} (-1)^{q-p} [z^{q-p}] (1+z)^{2n-2l} \\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} \sum_{p=0}^q {2l\choose p} (-1)^p z^p.$$
We may extend $p$ to infinity because with $p\gt q$ there is no contribution to $[z^q]$, getting
$$[z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} \sum_{p\ge 0} {2l\choose p} (-1)^p z^p \\ = [z^q] (-1)^q \sum_{l=0}^{n} {n\choose l}^2 (1+z)^{2n-2l} (1-z)^{2l} \\ = [z^q] (-1)^q [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n \\ = [z^q] [w^n] (1+w(1-z)^2)^n (1+w(1+z)^2)^n.$$
Re-write this as
$$[z^q] [w^n] ((w(1+z^2)+1)^2 - 4 w^2 z^2)^n \\ = [z^q] [w^n] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} w^{2p} z^{2p} (w(1+z^2)+1)^{2n-2p} \\ = [z^q] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{2p} [w^{n-2p}] (w(1+z^2)+1)^{2n-2p} \\ = [z^q] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{2p} {2n-2p\choose n-2p} (1+z^2)^{n-2p}.$$
We observe at this point that we get zero here when $q$ is odd, which agrees with the target formula. We are thus justified in putting $q=2l$ to get
$$[z^l] \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} z^{p} {2n-2p\choose n-2p} (1+z)^{n-2p} \\ = \sum_{p=0}^n {n\choose p} (-1)^p 2^{2p} {2n-2p\choose n-2p} {n-2p\choose l-p}.$$
Note that
$${n\choose p} {2n-2p\choose n-2p} {n-2p\choose l-p} = \frac{(2n-2p)!}{p! \times (n-p)! \times (l-p)! \times (n-l-p)!} \\ = {l\choose p} \frac{(2n-2p)!}{(n-p)! \times l! \times (n-l-p)!} = {l\choose p} {2n-2p\choose n-p} {n-p\choose l}.$$
Re-indexing we get for the sum
$$(-1)^n 2^{2n} \sum_{p=0}^n {l\choose n-p} {2p\choose p} {p\choose l} (-1)^p 2^{-2p} \\ = (-1)^n 2^{2n} \sum_{p=0}^n {2p\choose p} (-1)^p 2^{-2p} [z^{n-p}] (1+z)^l [w^l] (1+w)^p \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \sum_{p=0}^n {2p\choose p} (-1)^p 2^{-2p} z^p (1+w)^p.$$
We may once more extend $p$ to infinity because there is no contribution from the sum term to the coefficient extractor $[z^n]$ when $p\gt n,$ obtaining
$$(-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \sum_{p\ge 0} {2p\choose p} (-1)^p 2^{-2p} z^p (1+w)^p \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \frac{1}{\sqrt{1+z(1+w)}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^l [w^l] \frac{1}{\sqrt{1+z+wz}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} [w^l] \frac{1}{\sqrt{1+wz/(1+z)}} \\ = (-1)^n 2^{2n} [z^n] (1+z)^{l-1/2} {2l\choose l} (-1)^l 2^{-2l} z^l \frac{1}{(1+z)^l} \\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l} [z^{n-l}] \frac{1}{\sqrt{1+z}} \\ = (-1)^{n-l} 2^{2n-2l} {2l\choose l} {2n-2l\choose n-l} (-1)^{n-l} 2^{-(2n-2l)} \\ = {2l\choose l} {2n-2l\choose n-l}.$$
This is the claim. Credit goes to the Egorychev method which was presented here in formal power series notation.
Alternate answer. We keep the preliminaries and start with the contribution from $w$ being
$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w(1-z)^2)^n (1+w(1+z)^2)^n.$$
Now put $v=w/(1+w(1-z)^2)$ so that $w=v/(1-v(1-z)^2)$ and $dw = 1/(1-v(1-z)^2)^2 \; dv$ to get
$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} (1-v(1-z)^2) \frac{(1+4vz)^n}{(1-v(1-z)^2)^n} \frac{1}{(1-v(1-z)^2)^2} \\ = \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \frac{(1+4vz)^n}{(1-v(1-z)^2)^{n+1}} \\ = \sum_{p=0}^n {n+p\choose n} (1-z)^{2p} 4^{n-p} z^{n-p} {n\choose p}.$$
Observe that
$${n\choose p} {n+p\choose n} = \frac{(n+p)!}{p! \times (n-p)! \times p!} = {2p\choose p} {n+p\choose n-p}.$$
We obtain
$$(-1)^n \sum_{p=0}^n {2p\choose p} {-2p-1\choose n-p} (-1)^p (1-z)^{2p} 4^{n-p} z^{n-p} \\ = (-1)^n [w^n] \frac{1}{1+w} \sum_{p\ge 0} {2p\choose p} (-1)^p (1-z)^{2p} 4^{n-p} z^{n-p} (1+w)^{-2p} w^p.$$
Here we have extended to infinity due to the coefficient extractor in $w.$ Continuing,
$$(-1)^n 4^n z^n [w^n] \frac{1}{1+w} \sum_{p\ge 0} {2p\choose p} (-1)^p (1-z)^{2p} 4^{-p} z^{-p} (1+w)^{-2p} w^p \\ = (-1)^n 4^n z^n [w^n] \frac{1}{1+w} \frac{1}{\sqrt{1+(1-z)^2 w /(1+w)^2 /z}} \\ = (-1)^n 4^n [w^n] \frac{1}{1+wz} \frac{1}{\sqrt{1+(1-z)^2 w /(1+wz)^2}} \\ = (-1)^n 4^n [w^n] \frac{1}{\sqrt{(1+wz)^2+(1-z)^2 w}} \\ = (-1)^n 4^n [w^n] \frac{1}{\sqrt{1+w}} \frac{1}{\sqrt{1+wz^2}}.$$
Now here we see that there are no odd order coefficients in $z$ and we may put $q=2l$. Extracting the coefficient we get
$$[z^{2l}] (-1)^n 4^n [w^n] \frac{1}{\sqrt{1+w}} \frac{1}{\sqrt{1+wz^2}} \\ = [z^l] (-1)^n [w^n] \frac{1}{\sqrt{1+4w}} \frac{1}{\sqrt{1+4wz}} \\ = (-1)^n [w^n] \frac{1}{\sqrt{1+4w}} {2l\choose l} (-1)^l w^l \\ = (-1)^n {2l\choose l} (-1)^l {2n-2l\choose n-l} (-1)^{n-l} \\ = {2l\choose l} {2n-2l\choose n-l}.$$
This is the claim.