Which module shows the ones

The rule

Each byte you send pushes the ones before it further along. So after the last shiftOut, the byte that is still sitting in the module nearest your board is the one you sent last.

That module is on the right. The board's IN edge faces right when the digits face you, so every module you add goes to its left, and a number written left to right lands correctly — provided the first byte out is the leftmost digit.

Send 73 with buf[0] holding the 7 and it reads 73. Switch the second control and put the ones digit in buf[0] instead: the same hardware, the same wires, and the display reads 37. Nothing is miswired. The 3 simply left the pin first and therefore travelled furthest.

In the loop

That is why the sketch's digit-peeling loop counts down:

for (int i = NUM_DIGITS - 1; i >= 0; i--) {
  b[i] = d[t % 10];   // the rightmost digit of what is left
  t /= 10;            // drop it
}

t % 10 gives the ones digit, and it goes into the last slot of the array — because the last slot is sent last, and the last byte sent is the one that stays nearest your board.

Written the other way round it would still compile, still run, and still light every segment. It would just print the number backwards.

Sanity check before you write a clock

Send a number whose digits are all different — 12, or 123 — rather than 11 or 88. Two identical digits will read correctly whichever way round the array is filled, which is a fine way to convince yourself the wiring is right and then discover otherwise an hour later.