Trying unsuccessfully to heat a 12 V 1 A jacket with an 18 V battery and 5 ohm potentiometer
There are two reasons why your approach is off: Loading Effects and Series Resistance in the battery itself. Both will throw off your calculations if you don't consider them.

By connecting the circuit to the device you load the battery down (current flows out of it). Previously with the potentiometer not connected to anything, current was flowing already, you can measure this with a multimeter if you don't believe me. By loading down the circuit even more current is required then, so you are getting an even bigger voltage drop on the potentiometer because more current is going through it now. The voltage that's left is what goes to your device.

The effect of the resistance within the battery will cause your voltage to be lower than expected when high currents are required in your circuit. These high currents must leave the battery and cause a voltage drop at the battery terminals. It's not as big of an issue as loading effects but it's something to be aware of.

Here is a more accurate model including these facts. (Ignore the resistor values)







simulate this circuit &ndash; Schematic created using CircuitLab

You will also come to learn that a potentiometer is a bad design for dropping to a specific voltage! It's basically the reason you are experiencing loading effects in your circuit.

A better design would use a voltage regulator such as the LM7812.

However in the meantime, and perhaps the shortest answer to this question is: Tune the potentiometer output voltage to 12V while it's connected to the jacket.

Another fun thing that might happen is this voltage may drift as the wires in the jacket heat up and the effective resistance of the jacket changes. This is a discussion for another day.