Why is the "square of the deviation" always the tail end in the near-50 and near-100 methods?

Because a binomial square (base ± d)² ends in the d² term, which occupies the low-order digits.

Expand (b ± d)² = b² ± 2bd + d². When the base b is 50 or 100, the b² ± 2bd part is a multiple of 100 (it forms the leading digits), and d² is what's left in the last two places. So the deviation's square is literally the ones-and-tens of the answer — provided d² fits in two digits, else its overflow carries up. That single identity is the engine behind every squaring trick here.