So say I'm powering only the red led of an rgb strip and want it at 50% brightness
So low side switching is when the 12v is constant and the ground is switched on and off at 50% meaning 50% of the time its powered at 12v and 50% of the time the ground is disconnected and the circuit is interupted
Up to here, everything is fine.
and PWM where 12v would be modulated between 12-0v say 50% powering it at 12v half the time and powered at 0v half the time, while the ground is constant and the circuit is uninterrupted?
I don't understand that sentence, but from the following, i think it is wrong.
So (and I'm guessing here) with PWM a buck converter would be able to "smooth" the input voltage with capacitors to a constant non PWM voltage while with low side switching it would not have an effect since the circuit is completely interrupted?
Are PWM and high side switching the same or just similar?
No and no.
High and low side switching is just where you interrupt the circuit: Between + and load, or between - and load. This has no influence on the load. A simple load will not even be able to know how it is switches. PWM will remain.
For mechanical switching, usually the connection to plus is interrupted. This has multiple reasons, one of them is that often, a lot of stuff is at GND/- potential, so cutting of plus decreases the danger of shortcircuits, e.g. with the chassis. Cutting off phase in AC makes touching e.g. a lamp socket safe when switched off.
However, in electronics, there are multiple reasons for low side switching. One is that if you want non inverted drive, you need an NPN/n-channel transistor. An n-channel MOSFET has to compare the gate voltage to ground, so it has to be CONNECTED to ground. At the LOW side, the MOSFET source is connected GND and will have GND potential. At the HIGH side, source has + potential (minus a little coltage drop over the MOSFET), so the gate-source voltage is very small, or even negative. So that type of transistor would not work in that place. A PNP/NPN transistor would work, but had inverted gate logic.
The more important reason, however is, that you can freely choose your supply voltage, because your switching component only has to sink current, not provide it and thus have a certain voltage.
I think you mean power is defined by current and voltage P=I*V so at 12v 2A my header is only rated for 24 watts so there is the possibility exceeding the wattage WILL burn out the header
My PSU itself (as printed on the side) can supply the 5v rail with 15Amps
Since I'm using the header as ground and not using the power I don't need to worry about the wattage?
I don't see why a header should be rated for power. That does not make sense. Sending 10A over the header without any other load will require little voltage and not that much power, but it will probably melt the header. 220V @ .5A which drop somewhere else in the circut and deliver 110W on the other hand will probably be fine (but it is of course not advised to used those headers for mains).
Anyway, those 15A are meaningless, as long as you do not produce a short.