@wilriker
The wire on the heated bed pcb (or any other heater element) are behaving like resistors. If you are supplying power to a resistor, it heats up. Lets see some examples for a standard 12V heated bed:
(looks like this:)
A



B
U = 12V
R = 1 Ohm (This is a standard resistance for a 12V bed).
I = U / R ==> 12 / 1 = 12 Amp of current flows threw the bed.
P = U * I ==> 12V * 12A = 144W heat is being generated by our bed.
Let's see the same example with 24V supplied:
U = 24V
R = 1 Ohm (This is a standard resistance for a 12V bed).
I = U / R ==> 24 / 1 = 24 Amp of current flows threw the bed.
P = U * I ==> 24V * 24A = 576W heat is being generated by our bed, which is quadruple.
Now let's say, that the bed is actually 4 Ohm of resistance (this is a sandard 24V bed), but you put a third shoulder pad in the middle:
A

C

B
The resistance between A and C is 2 ohms, and B and C is 2 ohms aswell.
If you connect A and B to + and C to , then now you connected the two resistances (ac and cb) in parallel, so their resistance becames the half of each, so 1 ohms. 1 Ohm, 12V = 144W (see the calculation above)
But of you connect A to + and B to  (and leave C unconnected) then you conected the resistances in serial, so their resistance adds up to 4 Ohm:
Let's see the same example with 24V supplied:
U = 24V
R = 4 Ohm
I = U / R ==> 24 / 4 = 6 Amp of current flows threw the bed.
P = U * I ==> 24V * 6A = 144W heat is being generated by our bed, which is the same as using the bed from 12V and connecting the resistances in parallel, So now your bed can be used in either 12 or 24V and generates the same 144W, you just have to rewire it to your needs.