Manually selecting the current through the 1 W amber LED with a switch actually intended for LED flashlights:
The resistors on the low side of the LED use the MP1584 regulator for current control, with the orange wire feeding the resistor voltage into the error amplifier.
The 15 Ω unswitched resistor sets the LED current at 53 mA = 0.8 V / 15 Ω, with the LED dissipating about 100 mW. The resistor dissipates 43 mW.
Closing the switch puts the two parallel 4.7 Ω resistors in parallel with the 15 Ω resistor to produce 2.0 Ω, which sets the LED current to 390 mA and runs it at 950 mW. Each of the 4.7 Ω resistors dissipates 140 mW.
That much power raises the aluminum body to 50 °C = 120 °F: definitely uncomfortable but probably survivable for the LED inside.
Eyeballometrically, a decimal order of magnitude difference in the LED current produces an obvious brightness difference. My first try ran the LED at 500 mW (a binary order of magnitude less than 1 W) and wasn’t visually different. Given that the LED will run from the Bafang’s headlight output, saving power isn’t all that important.
If this is the first time you’ve encountered parallel resistors, this is why your calculator has a reciprocal button: the total resistance is the reciprocal of the sum of the reciprocals of all the resistances:
1/R = 1/R₁ + 1/R₂ + …