"0" volts generally means "tied to ground." Think of it as a potentiometer with one side tied to 10 V and the other tied to 0 V, as the wiper of the pot moves toward the 0 V terminal. (0 V is not necessarily "ground" although it can be; terminology again!)
This is what is confusing me, you can use a potentiometer as a dimmer for this driver. Where at 0 ohms it is a shorted and at high ohms essentially open. I am more confused about the 0-10v dimming aspect because that requires an input of 0-10v on the wires. If you can just use a potentiometer as a dimmer on the input wires then there has to be some output voltage(on the "input" wires) right? If so, how can you have 0-10v of input on wires that are already providing an output? Maybe I am just confusing myself or there is something which is flying right past me.
The dim input has 2mA constant current source connected to it, with an open circuit voltage of around 10V. The LED brightness depends on the average voltage on the dim input.
The dim input can be controlled by a voltage, resistance and quite often PWM, since it's the average voltage, which determines the brightness. If a low impedance voltage source, capable of 2mA is connected to the dim input, the brightness will simply change, with the applied voltage.
Ohm's law states the voltage is proportional to the resistance, so connecting a 500R potentiometer to it, will give a voltage range of near zero to 0.002*500 = 10V.
This is very common feature with LED drivers.