Why it needs 12 volts

What a segment's current passes through

On this board every segment is the same short path. The display's two anode pins are one wire, the common anode, and it is fed from the 12 V supply (through a switch, Q2, below). From there a segment's current goes through the LEDs inside the bar, through a resistor of its own (75 ohm for segments A to G, 510 ohm for the point), into one channel of U2, and from U2 to GND. U2 is the driver chip: it is the thing that switches the current, and a high on its input is what turns a segment on.

The display's own LED voltage was not measured, so the figure assumes one and says so: pick one LED or a string of four, and watch what is left for the resistor. What the board does tell us is the other two numbers. U2 uses about 1 V when it is on. And 75 ohm is a small resistor: at any current in the tens of milliamps it drops only a volt or two (20 mA across 75 ohm is 1.5 V). So with 1 V in the driver and a volt or two in the resistor, most of a 12 V rail must be sitting across the display itself. That is a lot more than one LED needs, and it is what several LEDs in a row in each bar would add up to.

That is the argument for the 12 V input, and it is an inference from those numbers rather than a reading. What holds either way: below the total a series string does not glow faintly, it does nothing, and a 5 V rail is under any total that has several LEDs in it.

The chip that decides is not the chip that drives

U1, the 74HC595, decides which segments are on. The datasheet rates it for a 2 to 6 V supply and about 6 mA out of one pin, which is an indicator LED on a breadboard and not a 4-inch bar.

So U1's eight outputs go to the eight inputs of U2, an ULN2803A. Each of its channels is a Darlington pair of transistors that sinks current to GND when its input is high, and each can take 500 mA absolute maximum. It needs at most 2.4 V on an input to sink 200 mA, so a 3.3 V high from U1 is enough. Because it sinks current and the segments' other end is the common anode, a 1 in your byte switches a segment on, which is what the table in the sketch assumes.

Three connections, and one of them bites

Twelve volts and its ground on the two-pin header. Your board's 3.3 V or 5 V and its ground on the signal header. And those two grounds must be the same wire.

The 12 V does not go straight to the anode. It goes through Q2, a MOSFET that acts as a switch, and that switch is how the board dims. The drawing assumes it is on; how it is driven is in how the dimmer works, and what it does to a module you have not wired for PWM is in the PWM pad.

Leave out the 12 V and everything else works perfectly: bytes arrive, the latch fires, and the digit is dark. Leave out the logic rail and U1 cannot store the byte. Leave out the shared ground and you get the bad one, a circuit that half works, because a HIGH is a voltage measured against ground and the two halves no longer agree where zero is.

What to buy

Any 12 V supply. Barrel jacks, screw terminals and bare leads are all fine; what matters is that its negative lead ends up on the same ground as your microcontroller, which usually means a wire from the supply's ground to a GND pin on your board as well as to the module. The back of the board prints a suggestion for getting 12 V from a USB-C power source, which is the one to read if you do not have a 12 V adapter.