I have a question about the operation of a voltage regulator. The supply is 12V and can provide 15W (max current draw 1.25A). Now when I use a linear voltage regulator down to 5V do the 15W still apply (3A current draw)? Or do the 1.25A from the power source still apply since no current is being transformed?
You're using a linear regulator which simply "burns off" the excess voltage.
The current does not change and remains the same so you can draw up to 1.25 A at the output of the regulator. So after the regulator you're limited to 5 V, 1.25 A so 6.25 W.
When you draw that 6.25 W there is 12 V - 5 V = 7 V at 1.25 A meaning 7 V * 1.25 A = 8.75 W dissipated in the regulator. It will get hot so use a heatsink! Note how 8.75 W + 6.25 W = 15 W which is the maximum power from the supply.
If you want more current at 5 V you will need to use a switched mode regulator (also called a "buck converter"). Then theoretically you could draw up to 15 W at the output so 5 V, 3 A. I write theoretically because this assumes no power is lost. In practice some power is lost and 5 V, 2.5 A would be a more realistic value. An LM2596 based DCDC converter board could do this job.
In a linear voltage regulator as the LM117, all the voltage drop × current is turned into heat. That's about 9W in you case. You can draw 1.25A@5V from the 5V output.
If you wanted to draw more current on the 5V side than it is supplied on the 12V side and produce less heat, you had to use a switching regulator. There are some manufacturers which produce drop-in replacements for the linear regulators of the LM78xx series. (google: 7805 switching)