While using a one-shot timer circuit will work, I think an easier solution can be used. Take a look at this circuit.
For clarification, "VBAT" is a 12V source that is always on as long as the battery is connected. However, "ACC" is a 12V source that is only on when the ignition is on or the key is set to "accessory." Rather than using a 5V relay just to control the power to the RPi, why not use a standard 12V auto relay as shown. This way, there is no wasted power (except for the coil current while the power is on) because everything will be disconnected from the battery.
One side of the coil is always connected to 12V. The opposite side is connected to ground (chassis) through an N-Channel FET (Q1). While a MOSFET is used in the diagram, any FET capable of sinking the coil current can be used. When "ACC" is powered ON, Q1 will switch ON, connecting the coil to ground and actuating the switch. This will in turn power whatever 5V regulation circuit you plan to use (a simple 7805 regulator with heat sink, a switching DC-DC converter, the USB supplies mentioned, etc).
The diode D2 is there to ensure the capacitor can only discharge into Q1 and can be regular or Shottky. Other methods should probably be used for over voltage and current protection from the battery.
The "ACC" voltage can be put through a voltage divider to create a 3.3V signal for the RPi. Be careful with this voltage level, considering a 12V auto battery can really be more like 14V DC. As long as this signal is HI, the RPi knows that the power is on. Obviously, this GPIO pin should be set as an input with any internal pullups disabled. When "ACC" is turned off, the RPi should see the LO signal on the pin and begin its shutdown.
When the "ACC" voltage is turned off, the capacitor C1 will retain the charge for so long, discharging through the resistor R1. Once the capacitor voltage drops below the gate threshold of Q1, it will switch OFF, disconnecting the relay coil from ground and removing power from the peripheral circuit. If a "logic level MOSFET" is used for Q1, it will remain switched ON until C1 voltage is fairly low. I tested this circuit using an NTD4960 (Datasheet), and it remained on for around 15 seconds - until C1 was around 2V. To increase the time, increase the capacitance value.
Best Answer
There may not be enough current available on the pins 2&4 "GPIO" (actually power pins). The Velleman relay should draw less than 100mA when on.
The +5V is drawn directly from the USB port with a resettable fuse in series. Looks like a 2A fuse on yours (which means it might trip at 1.5 or 1.3A at elevated temperature). The current is also limited by what the USB can supply. So if you look at those two limits and take the smaller you may get some idea.
There is insufficient information on the Velleman to determine if it will work reliably from a 0V/3.3V signal. No specification and no schematic.
I would guess it might work well enough for you but that's a guess. If you hook it up correctly it shouldn't hurt anything, but you may not be able to tell if it is marginal.