The identity $$\sum_k \binom{2 k}{k} \binom{2n - 2k}{n - k} = 4^n$$ is found on page 187 of "Concrete Mathematics" by Knuth and Graham. The book does not prove it combinatorially.
Is there a proof of this identity using double counting? I have played with it for about an hour to no avail!
Like I said, I don't know a direct proof of this, but I do know of one proof using binomial Taylor series. I'm pretty sure this proof is unelegant and overly complicated, but it works out.
The Taylor series for $\frac{1}{\sqrt{1-4x}}$ is given by $\sum_{n=0}^{\infty} {{2n}\choose{n}} x^n$.
Now recall the convolution identity for two series, assuming one of them converges absolutely:$$\bigg(\sum_i^{\infty} x_i\bigg)\bigg(\sum_j^{\infty}y_j\bigg) = \sum_n^{\infty} \bigg(\sum_k^n x_ky_{n-k}\bigg)$$ Applying this to the series for $\frac{1}{\sqrt{1-4x}}$ with itself, we have that $$\sum_{n=0}^{\infty} \bigg( \sum_{k=0}^n {{2k}\choose{k}}{{2(n-k)}\choose{n-k}} \bigg)x^n = \bigg( \sum_{n=0}^{\infty} {{2n}\choose{n}}x^n \bigg)^2 = \frac{1}{1-4x} = \sum_{n=0}^{\infty} 4^nx^n$$Then, using the fact that the terms in a series are uniquely defined, we get that $$\sum_{k=0}^n {{2k}\choose{k}}{{2(n-k)}\choose{n-k}}=4^n$$ as desired.
