Proving $\sum_{k=0}^n\sum_{l=0}^k{n \choose k}^2{k \choose l}{n \choose l}{2n-l \choose n}=\sum_{k=0}^n {n \choose k}^2{n+k \choose k}^2$

Here is one way (probably not the simplest). Looking at summands on both sides for fixed $k$, they do not match. But we can change the order of summations on the left so they do.

First we can notice that the inner sum on the left can actually go to $n$ since for $l>k$ we have $\binom{k}{l}=0$. Then, we can swap the summands, so we get left hand side $$ \sum_{l=0}^n\sum_{k=0}^n{n \choose k}^2{k \choose l}{n \choose l}{2n-l \choose n}. $$ Then by inspecting the sum for small values, we can see that individual summands now match the same values on the right side, just in the reverse order. So we revert the order by changing $l \mapsto n-l$.

$$ \sum_{l=0}^n\sum_{k=0}^n{n \choose k}^2{k \choose n-l}{n \choose n-l}{2n-(n-l) \choose n}. $$ If we also switch $l$ and $k$ to match right side, we get $$ \sum_{k=0}^n\sum_{l=0}^n{n \choose l}^2{l \choose n-k}{n \choose n-k}{2n-(n-k) \choose n}. $$ and we conjecture for fixed $n,k$: $$ \sum_{l=0}^n{n \choose l}^2{l \choose n-k}{n \choose n-k}{2n-(n-k) \choose n}={n \choose k}^2{n+k \choose k}^2, $$ which would imply your equality. But notice that already two terms are present on both sides since $\binom{n}{n-k}=\binom{n}{k}$ and $\binom{2n-(n-k)}{n}=\binom{n+k}{n}=\binom{n+k}{k}$, and so the equality is equivalent to $$ \sum_{l=0}^n{n \choose l}^2{l \choose n-k}={n \choose k}{n+k \choose k}. $$ The last equality is much simpler, I suggest to look at some proofs in Proving $\sum_{m=0}^n\binom{n}{m}^2 \binom{m}{n-k}=\binom{n}{k}\binom{n+k}{k}$ or Some binomial coefficient identity.

We seek to verify that

$$\sum_{k=0}^n {n\choose k}^2 \sum_{l=0}^k {k\choose l} {n\choose l} {2n-l\choose n} = \sum_{k=0}^n {n\choose k}^2 {n+k\choose k}^2.$$

Starting with the inner term on the LHS we find

$$\sum_{l=0}^k {k\choose k-l} {n\choose l} {2n-l\choose n} \\ = [z^k] (1+z)^k \sum_{l=0}^k z^l {n\choose l} {2n-l\choose n} \\ = [z^k] (1+z)^k [w^n] (1+w)^{2n} \sum_{l=0}^k z^l {n\choose l} (1+w)^{-l}.$$

The coefficient extractor $[z^k]$ enforces the upper limit of the sum and we find

$$[z^k] (1+z)^k [w^n] (1+w)^{2n} \sum_{l\ge 0} z^l {n\choose l} (1+w)^{-l} \\ = [z^k] (1+z)^k [w^n] (1+w)^{2n} \left(1+\frac{z}{1+w}\right)^n \\ = [z^k] (1+z)^k [w^n] (1+w)^{n} (1+w+z)^n.$$

We get from the outer sum

$$\sum_{k=0}^n {n\choose k}^2 [z^k] (1+z)^k [w^n] (1+w)^{n} (1+w+z)^n \\ = \sum_{k=0}^n {n\choose k}^2 [z^n] z^k (1+z)^{n-k} [w^n] (1+w)^{n} (1+w+z)^n \\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n \sum_{k=0}^n {n\choose k}^2 z^k (1+z)^{-k} \\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n \sum_{k=0}^n {n\choose k} v^k z^k (1+z)^{-k} \\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n \left(1+\frac{vz}{1+z}\right)^n \\ = [z^n] [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n (1+z+vz)^n.$$

Extracting the coefficient on $[z^n]$ we find

$$\sum_{k=0}^n ( [z^{n-k}] [w^n] (1+w)^n (1+w+z)^n ) ( [z^k] [v^n] (1+v)^n (1+z(1+v))^n ) \\ = \sum_{k=0}^n \left({n\choose n-k} [w^n] (1+w)^{n+k}\right) \left({n\choose k} [v^n] (1+v)^{n+k}\right) \\ = \sum_{k=0}^n {n\choose k}^2 {n+k\choose n}^2.$$

This is the claim.