In a typical DC/DC converter a proportion of the output voltage is fed back through a voltage divider into the regulator IC, which compares it to an internal reference voltage and raises or lowers the output voltage to make the feedback and reference voltages equal.
The block diagram looks something like this:-
simulate this circuit – Schematic created using CircuitLab
The effect of regulation is to maintain a constant voltage across R2 equal to the reference voltage (eg. 0.9 V), which Ohm's Law tells us must then have a constant current flowing through it (eg. 0.9 V / 1.2 kΩ = 0.75 mA). This current also flows through trimpot R1, so the voltage across it (and therefore the output voltage) will be directly proportional to its resistance (eg. Vout = 0.9 V + 0.75 V per kΩ of R1).
If we wire another pot in parallel across R1 the total resistance will be lowered and the output voltage will be reduced. The new pot can also have a resistor wired in series to set the minimum total resistance. The voltage divider circuit then looks like this:-
simulate this circuit
Everything inside the box is effectively a single resistor as far as the feedback circuit is concerned. Setting R4 to its maximum value and adjusting R1 sets the maximum resistance of the combination, thus setting the maximum regulated output voltage. Setting R4 to 0 Ω puts 6.8 kΩ across R1 which sets the minimum output voltage. Intermediate settings of R4 cause the output voltage to vary between these minimum and maximum voltages.
why a 22K potentiometer and would any other value work? And why would
a 6.8K resistor result in a minimum voltage of 5V?
The values of 22k and 6.8k were chosen (possibly by experiment) to suit the particular DC/DC converter. Using different values would result in a different voltage range. If your converter is not identical to the one in the video then you may need to experiment to determine the correct values for your device.