One latch per digit

What chaining needs

A 74HC595 has a ninth output, Q7S (QH' in TI's naming), pin 9. It is the bit falling off the end of the shift register: after eight more clocks, the byte that was inside comes out of it. Wire it to a second 595's DATA, share CLOCK and LATCH, and sixteen clocks fill both chips.

On the TK53, pin 9 is in no net. No trace leaves it and no header pin carries it. The page this book replaces said the board chains; it cannot.

What works instead

Wire both boards' DATA to one pin and both CLOCKs to another. Every byte you shift goes into both shift registers. Give each board its own LATCH, and raise only one: that board copies the byte to its digit, and the other keeps what it had.

void showOn(int latchPin, byte pattern) {
  digitalWrite(latchPin, LOW);
  shiftOut(DATA_PIN, CLOCK_PIN, MSBFIRST, pattern);
  digitalWrite(latchPin, HIGH);
}

Two digits cost four pins, three cost five. A chain would cost three for any number. For more than two or three digits, a four-digit display with a driver chip of its own, such as the TK51, is less wiring.

Every board draws its own segment current from VCC, so two 8s at 3.3 V are up to 140 mA from your board's 3V3 pin. Most ESP32 boards' regulators supply that comfortably; add up the current before adding a fourth.