One possible circuit is shown below (one for each side)
TR1 and TR2 (PNP Darlington pair) operate as a switch that connect the indicator to running power via resistor R2. R2 is chosen so that lamp is at half power so may not be 12 ohm.
When indicator power is applied, R2 is bypassed by diode D2 so that lamp is more or less at full power. At the same time C1 gets fully charged. The high voltage on base of TR2 turns it off, turning off TR1.
When indicator power is turned off, capacitor C1 discharges via resistor R1. This causes the voltage on the base of TR2 to drop until it is low enough to turn TR2/TR1 on so the lamp, which was off, is again at half power.
The value of R1 and C1 are chosen to be a little bit more than the indicator 'off' time.
When indicator is off, the lamp runs at 1/2 power.
When indicator is on, the sequence is full power, off, full power, off ......
When indicator is turned off, after the last lamp off sequence, the indicator returns to run at 1/2 power after the time determined by C1/R1. You will need to play with value of C1 and R1.
An approximate analogy using water to represent electricity is below.
The outlet pipe (R1) empties the bucket (C1). Increasing the diameter of the outlet pipe is equivalent to lowering the value of R1. Increasing the size of the bucket is equivalent to increasing the value of C1. Both of these factors determine how quickly the bucket is emptied. Inside the bucket is a float on a pivot that operates a switch (TR1/TR2) when the float drops to a specific level.
The bucket is constantly being emptied but when indicator power is applied, it gets filled instantly. Once indicator power is removed, the water level (voltage on C1) starts to drop and eventually the switch is operated so the lamp, which was off, runs at half power.