24v in 12 out

The resistor you use to reduce the voltage from 24 Volts to 12 Volts depends on the current required by the headlights. You need to know the current rating or power rating of the headlights. The resistor you have for reducing 12V to 6V probably is not the right one to use for 24V to 12V.

You can get a suitable resistor from a local electronics store or an on-line electronics retailer like Digikey or Mouser.

Here is an example of how to determine what resistor you need...

EXAMPLE
if the headlights have a rating label that says "12 Volts, 24 Watts" and they are connected in parallel (both pairs of wires are connected to the voltage source instead of in a series chain) the total power is 48 Watts. And 12V applied to a 48W load needs to provide 48W/12V = 4 Amps.
To get 12V from the 24V battery you need the resistor between the battery and headlights (assuming one resistor for two headlights). In this example, the resistance of the resistor needs to cause a voltage drop of 12V at a current of 4A because the headlights require 4 Amps of current at 24 minus 12 volts.
The resistance required is 12V/4A = 3 Ohms.
You also have to worry about the power dissipated in the resistor as heat and choose a suitable power rating. 4A x 12V = 48W dissipated in the resistor. Choosing a resistor rated for 75W to 100W would provide some margin for hot days, poor heat conduction, or poor air circulation around the resistor. The nominal power rating of a resistor is only valid under certain conditions specified by the manufacturer, such as operation at room temperature and mounting to a metal plate. For worse conditions, you need a higher power rating than calculated.

For your particular case, you can use the following calculations:
RESISTANCE REQUIRED
R = (Vh/Ph)*(24-Vh)
where R is the resistance
Vh is the rated voltage of the headlights
Ph is the rated power of the headlights

RESISTOR POWER DISSIPATION
Pr = (Ph/Vh)*(24-Vh)
where Pr is the rated power of the resistor
Ph is the rated power of the headlights
Vh is the rated voltage of the headlights

The above formulas come from these basic electrical equations:
CURRENT
I = Ph/Vh
where I is the total current through both headlights (which is the same as the current through the resistor)
Ph is the total power rating of the headlights (or the total power used by both headlights at the rated voltage)
Vh is the voltage rating of the headlights

VOLTAGE
Vr = V - Vh
where Vr is the voltage across the resistor to reduce the battery voltage down to the rated voltage of the headlights
V is the battery voltage (24V in your case)
Vh is the voltage rating of the headlights (12V in your case)

POWER
Pr = I x Vr
where Pr is the power dissipated in the resistor
I is the same total current as above (all the current flowing through the headlights also flows through the resistor)
Vr is the voltage across the resistor calculated above

Final comments..
The best solution wold be to get 24V headlights and avoid the resistor altogether. With 12V headlights, the same amount of power used by the headlights is also getting wasted as heat in the resistor. Any increase in the rated voltage of the headlights would result in less heat and wasted energy in the resistor.

Good luck with solving your problem. I am interested in hearing about the results!