After $f$ folds the strip is $2^{10-f}$ squares long and $2^f$ layers thick, so the positions $L$ from the left and $R$ from the right satisfy $L + R = 2^{10-f} + 1,$ and the positions $B$ from the bottom and $T$ from the top satisfy $B + T = 2^f + 1.$ When the right half is folded over onto the left, a square in the left half keeps its $L$ and $B,$ while a square in the right half is flipped: its new $L$ is its old $R,$ and its new $T$ is its old $B.$

The $942$nd square starts at $(L, B) = (942, 1).$ Applying the rule through the ten folds gives $$\begin{gathered} (83, 2),\ (83, 2),\ (83, 2),\ (46, 15),\\ \ (19, 18),\ (14, 47),\ (3, 82),\\ \ (3, 82),\ (2, 431),\ (1, 594). \end{gathered}$$ For example, at the fourth fold the strip has length $128$ and $L = 83 \gt 64,$ so the new $L$ is $128 + 1 - 83 = 46$ and the new $T$ is the old $B = 2,$ making $B = 16 + 1 - 2 = 15.$

In the final stack of $1024$ squares, this square sits at height $594$ from the bottom, so $594 - 1 = 593$ squares lie below it.