Here's a drawing of what you want to do, though you may use a different number of LEDs. If the resistors are the same value then the current will be approximately double when you hit the brakes. Assume you want 30 mA when you hit the brakes and 15 mA normally. The LED datasheet (N46AT) shows that typical forward voltage = 1.8 V when current = 15 mA (see the I-V graph on the last page). If your LEDs are typical, then 6 in series would make 10.8 V at 15 mA. Not sure what type of battery you're using but assume it's a lead acid, so about 12.6 V when full. That gives 1.8 V left over to drop across the resistor. You must choose the resistor that will give 15 mA with 1.8 V across it. 1.8 / .015 = 120 ohms. If the other resistor is also 120 ohms, you'll have 60 ohms equivalent (two 120's in parallel) when you activate the contact switch. 1.8 / 60 = 30 mA.
Now if you hook up your battery right after charging, the voltage will be higher and you'll get higher currents. So do the above calculation using the highest voltage you'll connect to the circuit. You want to be sure the LEDs don't see more than 30 mA with a full battery. As the battery drains of course the current will decrease and your lights will dim.
If your LEDs don't follow the typical characteristics you'll have to play around with the resistor value. Start high so you don't end up burning up the LEDs. Good luck