We claim that $\boxed{\text{Jesse}}$ can always win. Let Jesse place $2^0 = 1$ in the black box. In each of his moves, Jesse either places the number half the smallest power of two, or the number twice the largest power of two which he has already placed. For example, if Jesse has already placed $2^{-i}, 2^{-i+1},\dots,2^{-1}, 2^0, 2^1, \dots, 2^{j-1}, 2^j$, then he could either place $2^{-i-1}$ or $2^{j+1}$ on his next turn. The box in which Jesse would place his number depends on Tjeerd's actions. The goal of Jesse is to keep the largest power of $2$ is the black box after his turn. If Tjeerd transfers the largest power of $2$ to the white box, then Jesse places the number twice of that in the black box, keeping the largest power of $2$ in the black box. If Tjeerd does nothing or transfer a power of $2$ which isn't the largest, then Jesse places the number half the smallest power of $2$ in any box available, keeping the largest power of $2$ in the black box. Note that Jesse has the final move, so at the end of the game the largest power of $2$ is in the black box. Since $2^{-i} + 2^{-i+1}+\dots+2^{-1}+ 2^0+ 2^1+ \dots+ 2^{j-1}< 2^j$, Jesse wins.