Why k-lists generalized birthday problem when $k=2$ is classical birthday problem?
They are essentially the same, certainly in terms of complexity. Any collision $x=y$ with $x\in L_x$ and $y\in L_y$ is a collision in a single list $L_x \cup L_y$ of size at most twice the size of the larger list.

Any collision in a single list is compatible with about half the partitions of that list into two equal sized halves. So an algorithm that finds a collision in the union randomly has found a collision in the two-list problem with probability $1/2.$

Moreover in the crypto context the lists contain (pseudo)random outputs of hashes or other crypto primitives.