How does a Low Power Factor draw more current?
However, something here just doesn't make sense. If we consider the power supply as an ideal voltage source, then the current has nothing to do with the given power.



The in-phase current has everything to do with the power. \$ P = VI \$.
The out-of-phase current does not contribute to the power but does contribute to the overall current drawn from the supply. The increased current causes an increased voltage drop along the lines giving poorer voltage regulation at the load. This is a bad thing!



  It seems that he just assumed the load would draw the required power to function properly no matter what.



It will draw the required power and this will be measured in kW and billed in kWh.
It will also draw reactive power measured in kVAr and billed in kVArh. Typically the utility will charge for excessive kVArh.



  On the other hand, it seems to me that the current would stay the same, being affected by the impedance of the load, leading to a lower real power and thus a malfunctional load.


No.



Links:

See my answer to How do capacitors improve power factor without destroying the capacitor?. It may help.



Simulation







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. A complicated struggle with CircuitLab's AC simulation capabilities.

Note that the values of L3, L4 and C2 have been chosen to have an impedance of 1 Ω at 50 Hz. The parallel R5-L3 pair should be out-of-phase by 45°. L4 and C2 should cancel out as their impedences are the same at 50 Hz. The 1 mΩ resistors are to help the simulator settle down after the initial transients.



Figure 2. The current waveforms for Figure 1. Note that the brown should be superimposed on the blue. The simulator is struggling.


The voltage plot has been omitted for clarity but the blue curve is in phase with it and will be taken as the reference.
The current for circuit (a), the resistor, is plotted in blue. As expected 1 V into 1 Ω gives 1 A peak.
The current for circuit (b), R5 and L3, is plotted in orange. Notice that the orange curve crosses the blue curve exactly at its peak (90° into the blue cycle). This occurs because the inductor current is lagging the voltage by 90° and is zero at this point. As the inductor current continues to rise the resistor current is falling. The peak occurs at 45° into the cycle where both currents are at \$ \frac {1}{\sqrt 2} I_{PEAK} \$. It follows that the peak of this waveform will be \$ \sqrt 2 \$ times that of the blue.
Finally note that if we add a capacitor of equal impedance to the inductor as in (c) the brown curve that we are back in phase again and the current has reduced. (I was expecting this to overlap the blue curve. I suspect the simulator is struggling.)


In summary:


The generator is constant (AC) voltage output.
The generator provides the current required by the load.
If the load is reactive (not purely resistive) the current increases while the useful power transferred remains the same.
By balancing the inductive and capacitive elements in the load the power factor can be brought back to unity.