12 VDC LED strips get too hot
It is almost certain that the string draws more than 120 mA from the transformer.
It is possible that you have misread the transformer current rating. This is dealt with below. 

Your string will almost certainly have multiple groups of 3 LEDs. Occasionally they are in one package but usually you get 3 LEDs from 12V to ground with a single common resistor per group. Occasionally you see groups with two resistors per group, presumably to spread the power dissipation amongst several resistors. So ...

If you can measure string current with car and adapter used, do so.
Also measure the voltage input at the connection point to the string in each case.
Go to "Jump to here if currents known" below 

Otherwise - voltage measurements needed. Any DMM (digital multimeter) will do.

Identify what constitutes a group. 

Count number of groups total.

Identify a series resistor.
Determine its value either from a marking on it or by measuring the resistance with an Ohmmeter with power turned off. 
[Typical values may be 5 Ohms to 200 Ohms. Range is wider than expected because they probably put several LEDS per package. This method handles that OK.]

Apply 12V from car.  

Measure voltage across the above resistor = Vr.
Measure voltage input to string at input point. Call this Vcar. Nominally 12V but may not be. 

Turn off power.  

Current in one group = V/R from above
eg = V = 3V and R = 20 Ohms then
Igroup = V/R = 3 / .020 = 150 mA.
Hopefully it will be rather less.

Total current = Groups x Igroup
Call this Icar

Repeat the exercise using the 12V supply. This will give you an idea of the current actually being drawn from the nominally 120 mA supply.
Call this Iadapter

Measure the voltage at the power supply connection point as it will be "loaded down" by the string. Call this Vadapter. If it was not loaded down the current would be the same as when connected to the car. 

Jump to here if currents known

Now:

The aim is to reduce string voltage to adapter equivalent. 
So you need to "drop" Vcar - Vadapter.
R = V/I = (Vcar-Vadapter) / I adapter
= resistor to use.

Power dissipation in resistor = I^2 x R = Iadapter^2 x R
Use a resistor with a somewhat higher wattage rating than this. Bigger than needed is OK.