Calculating voltage in a simple circuit
You're forgetting to include the forward voltage drop of the LED. The LED doesn't act like a resistor - it drops 2V relatively independently of the current through it (see the forward section of this graph). So the best way of describing the circuit is as having a 12V supply with 2V dropped by the LED and 10V across a 500 Ohm resistor, giving a circuit current of 20mA. Kirchoff's current law tells us that the current in the LED must be equal to the current in the resistor, so we're not missing anything out in our description. 

V=IR only works for resistive loads, and the LED part of the circuit is not resistive, as shown in the graph above - hence using the 2V forward drop model for the LED. If you wanted to replace the whole circuit with a resistor that drew the same amount of current at that voltage you could calculate this as 12V / 20mA = 600 Ohms - but that value would change if the supply voltage changes, as the LED does not exhibit a constant resistance of 100 Ohms, but instead is better modelled by the 2V drop model you are already using.

By the way, this isn't a daft mistake - understanding the use of different models for different circuit components and their combination is key to understanding basic circuit analysis. Learning is takes time and often feels counterintuitive to begin with. Hang in there - it gets better.