I checked my AC voltage at the transformer inputs and it measured at 121V. I can't do the math but maybe it was drawing too much.
All the matches are easy.
We have TE70023 transformer, aspect ratio of the single secondary winding in 115V connection (according to the datasheet) is 21.3V/115V = 0.1852, number of windings - 2, efficiency - 68%.
According to the Bugle power supply datasheet: full load is 2x15V @ 0.1A. On the output circuit we can find two bleeding resistors (R6, R9) drawing additional 2x (15V / 1000Ohm = 0.015A). Just before the regulators we can find another pair of such resistors (R5, R8) and marked voltages on them. They are drawing: 2x (20V / 10000Ohm = 0.002A). Until this point we have current 0.1A+0.015A+0.002A = 0.117A. There are two more just across the secondaries. The voltage across them could be counted as 21V + 0.117A*1Ohm=21.117V (because some voltage is across R1 and R11), and current as 2x (21.117V / 1000Ohm = 0.021). Total current from the transformer is 0.117A + 0.021A = 0.138A RMS.
The current on the primary windings would be 0.138A * 2 windings * 0.1852 = 0.051A = 51mA. There is question about efficiency though, because some current would be additionaly drawn to heat the transformer. The lost would be less than 100%-68%=32%, because we use not full sinusoudal period, but pulse current. Approximately it should be around 25%, so the resulting primary current should be 51mA / 0.75 = 68mA.

When your voltage is more than 115V (say 121V) we should add (121V/115V)*100%-100%= 5,22% to all voltages that are not regulated. So, the resulting secondary current would be:
0.1A + 0.015A + (0.002A * 1.0522) + (0.021A * 1.0522) = 0.139A RMS. Primary winding current should be then: 0.139A * 2 * 0.1852 / 0.75 = 68.6mA. Not very much of overhead!

I hope all has been counted right.

Not very much time just now to check.
My conclusion: 63mA fuse is for 230V usage.