Voltage regulator help
The wattage isn't something you inflict on a load but the voltage is. It seems to me that you might be taking the voltage (12V) and reading the label on the power supply (that says 10W) and therefore believing you will be putting 10 watts across to any load connected to it.

This isn't true - you put voltage on a load and the load determines what current is taken by ohms law (I = \$\dfrac{V}{R}\$).

So, if the load resistance is 14.4 ohms, the current will be 0.8333 amps and the power delivered will be 12V x 0.8333A = 10 watts.

If you inserted a 5V regulator between supply and 14.4 ohm load, the load would see 5 volts and take a current of 0.347 amps (ohm's law again) - this would be a load power of 1.736 watts. However, if the voltage regulator were a linear type there would be 7 volts dropped by the regulator at a current of 0.347 amps and this would be a power of 2.43 watts lost in the regulator.

Now the total power consumed is 2.43 watts + 1.736 watts = 4.17 watts (not 10 watts). It's load dependent.

If you used a near-100% efficient switching regulator there would be no power lost in the regulator and the power taken from the power supply would be just that consumed by the load i.e. 1.736 watts.