Pick a block size B = 2k−1. Somewhere in the vEB recursion is the level whose subtrees have height k = lg(B+1): those chunks are contiguous in memory — each is exactly one block — so a search reads only one per chunk. The shading below is that level, and it changes as you change B.
Height is fixed to h = k·2j so the recursion halves cleanly onto the height-k level and every B-piece is uniform (2k−1 = B nodes) and block-aligned — the cleanest case, using B = 2k−1 (not a power of two). The path has length h = log₂N but enters a new B-piece only every k levels, so it touches h/k = logBN chunks — that is the O(logBN) bound. For a general B each chunk still fits in ≤ 2 blocks.