Upper bound for set disjointness under product distributions
During the first phase of the algorithm, the players zoom in on a high-entropy coordinate. They exchange $O(1)$ bits, and stop at each step with probability $\Omega(\epsilon^2/\log^2(1/\epsilon))$ (here we use independence). Hence this phase uses up at most $O(\log^2(1/\epsilon)/\epsilon^2)$ bits in expectation. The second phase uses $O(n\epsilon)$ bits. In total, we used this many bits in expectation:
$$
O(\log^2 (1/\epsilon)/\epsilon^2 + \epsilon n).
$$
Choose $\epsilon = 1/n^{1/3}$ to obtain the stated bound.

Without independence, we don't have any control on the stopping probability at each step of the first phase. For example, it could be that $X$ is sampled uniformly at random, and $Y$ is the negation of $X$. In this case the first phase will never stop, and so the protocol will always use $2n$ bits.

As an aside, the upper bound can be improved to $O(\sqrt{n} \log n)$, which almost matches the lower bound $\Omega(\sqrt{n})$ (the gap might have been closed in the literature). See lecture notes of Prahladh Harsha.