Where the current goes

One loop

There is exactly one path on this board. SIGNAL goes to the LED's anode, the long leg. The cathode, the short leg, goes to the resistor. The resistor goes to GND. Your board closes the loop: its pin at one end, its GND at the other.

Current needs the whole loop. Break it anywhere, whether that is a missing GND wire, a loose jumper or SIGNAL on the wrong pin, and nothing flows anywhere in it. That is why the first check for a dark LED is always the ground.

HIGH

digitalWrite(pin, HIGH) connects the pin to the chip's supply, 5 V on an Uno. The LED takes about 3.0 V of that, which is what it costs to make blue light. The resistor takes the remaining 2.0 V, and 2.0 V across 150 Ω is about 13 mA.

That 13 mA comes out of your microcontroller's pin. The pin is sourcing current: pushing it out into the circuit. Every milliamp the LED uses, the pin supplies.

LOW

digitalWrite(pin, LOW) connects the pin to ground instead. Now both ends of the loop are at 0 V and there is no voltage to push current anywhere. The LED is off because nothing is flowing through it.

Why the board says ACTIVE HIGH

Active high is the name for "HIGH turns it on". The silkscreen prints it on the front, and GLOWS WHEN HIGH on the back, because not every board works this way.

Many dev boards wire their built-in LED the other way round: from the supply, through the LED, into the pin. On those, the pin has to go LOW to pull current through, and they light on LOW. That is active low, and it is why a sketch written for one board's built-in LED can look inverted on another.

This block cannot be active low. The resistor's far end is soldered to GND, so a voltage on SIGNAL is the only thing that can light it.