Challenge: How to prove this identity between bi- and trinomial coefficients?

We may prove this in a slick way via the Snake Oil method of Wilf (see link in the comments of the OP). To make notation far simpler, we note that the range of $k$ and $l$ is enforced by the binomial coefficients--they vanish for integers not valid for counting--and so sums over $k,l$ are taken over $\mathbb{Z}$.

That convention in mind, we multiply the RHS by $x^l$ and sum over all integer $l$:

\begin{align} \sum_{\ell,k} \binom{n}{\ell-2k,k,n-\ell+k}2^{\ell-2k} x^\ell &= \sum_{\ell, k} \binom{n}{\ell,k,n-\ell-k}2^{\ell} x^{\ell+2k}\\ &= \sum_{\ell, k} \binom{n}{\ell,k,n-\ell-k}(2y)^{\ell} (x^2)^k\\ &= (x^2+2x+1)^n \end{align}

In the first equality, we have shifted $\ell \mapsto \ell+2k$; rearranging slightly, we recognize the form of the trinomial theorem and thereby sum the series. But we can factor this expression to $(1+x)^{2n}$ and so apply the binomial theorem to obtain the $l$-th coefficient as $\binom{2n}{l}$. Comparing this with our first expression gives the desired identity.

Suppose we seek to evaluate (do some re-writing as observed by the OP) $$\sum_{k=0}^n {n\choose k} 2^{l-2k} {n-k\choose n-l+k}.$$

Start from $${n-k\choose n-l+k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-l+k+1}} (1+z)^{n-k} \; dz.$$

This gives the following integral for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^n {n\choose k} 2^{l-2k} \frac{1}{z^{n-l+k+1}} (1+z)^{n-k} \;dz \\ = \frac{2^l}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l+1}} \sum_{k=0}^n {n\choose k} 2^{-2k} \frac{1}{z^k(1+z)^k} \;dz \\ = \frac{2^l}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l+1}} \left(1 + \frac{1}{4z(z+1)}\right)^n \; dz \\ = \frac{2^l}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l+1}} \left(\frac{1+4z+4z^2}{4z(z+1)}\right)^n \; dz \\ = \frac{2^l}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l+1}} \frac{(1+2z)^{2n}}{2^{2n} (z(1+z))^n} \; dz \\ = \frac{2^{l-2n}}{2\pi i} \int_{|z|=\epsilon} \frac{(1+2z)^{2n}}{z^{2n-l+1}} \; dz.$$

This last integral can be evaluated by inspection and yields $$2^{l-2n} [z^{2n-l}] (1+2z)^{2n} = 2^{l-2n} \times {2n\choose 2n-l} 2^{2n-l} = {2n\choose l}.$$

A trace as to when this method appeared on MSE and by whom starts at this MSE link.