There is a basic problem when trying to apply Ohm's law to the analysis of a circuit with a rotating machine in it. The problem being the round thing with the shaft sticking out of it.
The current draw of a motor is determined by the voltage applied to it and THE MECHANICAL ENERGY DELIVERED TO THE LOAD by the shaft.
A motor running with NO mechanical load will draw a very small amperage. As you apply mechanical load to it, the amperage will increase rapidly as the motor slows down.
If we had a test motor and were able to have a variable speed 'load' attached to it, as you increased the speed of the variable speed 'load', you would see the current in the circuit actually drop to zero. If you continued to increase the speed, the current would reverse and our test motor would become a test generator. The key is MECHANICAL energy flow. As we speed up the 'load', we reverse the flow of mechanical energy from out of the test motor to into the test motor at the same time reversing the flow of current from into the motor to out of the motor.
Any motor that is turning either from the electrical energy supplied to its terminals OR from the mechanical energy supplied to its shaft is generating an internal voltage called a 'back EMF'. If the motor is being turned by the supplied voltage, then back EMF will tend to OPPOSE the supplied voltage.
In plain English, the motor behaves like a variable voltage generator that is directly tied to the speed the motor is turning. The voltage OPPOSES the voltage applied by the vehicle battery. The motor will accelerate until a balance is established between the applied voltage and the internal voltage at which point the motor will run at a steady speed.
If you apply twice as much voltage to the motor, the motor will have to turn twice as fast in order to reach this balance point.
If you stick a resistor in the supply circuit, then some of the supply voltage is 'dropped' across it and the motor runs at a slower speed.
Whats the current? That depends on the amount of mechanical power required to turn the fan, the speed of the motor, the battery voltage, the size of the voltage dropping resistor and a couple of other minor things.
The ceramic encased wire coil on the heater switch is a voltage dropping resistor used to provide slower speeds for the fan. It is normal for it to be VERY hot when running the fan at one of the slower speeds. In high fan position, there is normally not be a resistor in the circuit and the switch shouldn't be more than just warm on high speed.
