Here is a more detailed analysis of just the timer part. For the purpose of the simulation, I am omitting the various other switches and relay contacts required to latch the relay, and I am simulating only the timer details.
I am assuming that the relay has a 12Vdc coil with a resistance of 85 Ohms or higher. The diodes are 1N4007 Si rectifier types. I am adjusting the R1-C1 to get a relay drop-out time of about 10sec.
V2 is used to simulate the V(key) signal (ignition switch). In the sim, it is initially 0V (ign switch off), it becomes 12.8V at 2sec (ignition switch on), and back to 0V (ign sw off). Diode D1 conducts to charge C1 to the battery voltage (12.8V) minus one diode voltage drop. Note that V(RC) follows V(key) as it goes from 0V to 12.8V, but decays slowly after V(key) goes back low. To limit that charging current in-rush, it is necessary to add R2 (add this to the actual schematic you build).
After the ignition is turned off, the charge stored in C1 flows through R1 into the base of Q1. The base current is amplified by Q1 such that its collector current is sufficient to pull-in the armature of L1. For most relays, they drop out when their coil current becomes about 25% of their normal rated pull-in current. I plot I(L1), the current through the coil, and show where I think the relay drops out.
With the values shown for C1 and R1, it looks like that will produce a delay of about 9sec. To increase the delay, you can increase the value of C1. To shorten the time, you could make R1 less than 10K, but do not increase R1 above 10K to get longer delays. Obviously, if the ign sw is re-closed before the delay elapses, the relay never drops.