Shift, then latch

Two registers in one chip

A digit has eight LEDs: seven segments and a dot. Wired straight to a microcontroller that is eight pins. The 74HC595 does it with three, because it holds two rows of eight bits:

• the shift register, which fills one bit at a time from DATA, and
• the storage register, which drives the eight outputs and so the segments.

CLOCK fills the first. LATCH copies the first into the second.

Eight clocks

On each rising edge of CLOCK, the chip reads DATA, puts that bit into the first place, Q0, and moves every other bit one place towards Q7. The bit that was in Q7 falls off the end. After eight edges the eight bits you sent fill the register, the first one sent furthest along.

DATA is only read at the edge. Between edges it can change as it likes, which is how shiftOut works: set DATA, raise CLOCK, lower it, next bit.

One latch

All that time the digit shows the old byte. The outputs only change when LATCH goes from LOW to HIGH: then the whole shift register is copied to the outputs in one step. That is why a sketch writes

digitalWrite(LATCH_PIN, LOW);
shiftOut(DATA_PIN, CLOCK_PIN, MSBFIRST, pattern);
digitalWrite(LATCH_PIN, HIGH);

and why the digit goes from 1 to 2 without passing through anything in between. If LATCH were wired to CLOCK, the outputs would follow the register one edge behind and you would see each bit slide across the segments.

The chip is quick: Nexperia specifies a clock pulse of 15 ns at 4.5 V. No Arduino or MicroPython pin toggles that fast, so there is nothing to slow down.