Confused again. So if I had a 12V 0.5A (6W) power source, and I want a converted voltage of 6V. I will have my output at 6V 0.5A? or 6V 1A ?(Power=Voltage x Current = 6W)

The LM317 is a linear regulator. A characteristic of linear regulators is that CURRENT IN is the same as CURRENT OUT. The input and output voltages do not matter. It is how this type of regulator works.

Since current stays contant, the voltage difference from input to output is dropped across the regulator.

So if your load is drawing 500mA, this is how the math works.

Input: Vin * Iout = 12V * 0.5A = 6W

Output: Vout * Iout = 6V * 0.5A = 3W

Regulator's voltage drop: Vin - Vout = 12V - 6V = 6V

Regulator: Vreg * Iout = 6V * 0.5A = 3W

Now, consider if your input voltage was 9V and the output was 6V.

Input: Vin * Iout = 9V * 0.5A = 4.5W

Output: Vout * Iout = 6V * 0.5A = 3W

Regualtor's voltage drop: Vin - Vout = 9V - 6V = 3V

Regulator: Vreg * Iout = 3V * 0.5A = 1.5W

See, the total Power In is equal to the total Power Out PLUS the Power wasted by the regulator.

If you want to get more voltage or current from an input, you need a DC-to-DC boost converter. A simple linear regulator will not work that way.