A resistor running at its full power will get really hot so it is a good idea to have a resistor of a larger rating.
The resistor in parallel with the alternator warning lamp is needed to get the alternator started, and 100 Ω sounds good for that. It will dissipate around 1.5 W when on, so a 3 W resistor is a reasonable rating, and it will still get hot if left on.
If you have a 6 V LED, that will almost certainly be an LED with a resistor in series to allow it to run on 6 V. You should measure the current it takes at 6 V, and then you need a resistance of 6 divided by the current, and put that in series. So if it takes 20 mA you need 6/0.02 = 300 Ω.
(The 6 in the calculation is the extra voltage that you need to loose. It is calculated from 12 V (battery) - 6 V (LED) = 6 V (extra) and it just happens to be the same number as the LED voltage).