In a purely algebraic way, I've just stumbled onto the fact that

$$2^{2k} = \sum_{i+j=k} \binom{2i}{i}\binom{2j}{j},$$

i.e. that the self-convolution of the sequence $\binom{2k}{k}$ is the sequence $2^{2k}$. This is the type of identity for which I would expect there to be a beautiful, simple, direct, and probably well-known combinatorial argument. I just spent some time looking for it but couldn't find it.

Is there a direct combinatorial proof of the above?

Geometric remark: such an argument would partition the vertices of an even-dimensional hypercube into some sort of combinatorially meaningful classes of size $\binom{2i}{i}\binom{2j}{j}$. For example, for $k=2$, the vertices of the $4$-cube would be partitioned into classes of size $6=1\cdot 6$, $4=2\cdot 2$, and $6=6\cdot 1$.