Electrical – Wiring a 6 pin On/Off/On toggle switch (DC) 2 Pole

I don't quite understand the use of D4 to the Arduino Vcc pin. If it's intended as the primary Arduino power source you're going to have some problems because there's a rather large resistor (R6) in the way. \$\frac{12V}{10k\Omega} = 1.2mA\$. This is way too small for running the Arduino reliably. There's no need for it otherwise.

I would just hook the Arduino Vcc pin directly to the battery. Use a schottky diode if you intend to use multiple supplies, otherwise you can get rid of D4 completely.

There's also the issue with D1 and having the sense line hooked through the igniter. Personally I would rather not have any current flowing through the igniter unless I intend to fire it. If D1 is a fly-back diode, hook it directly around the igniter only.

Also, your voltage divider S2 sensor circuit is outputting a nominal 6V. This likely just barely in the acceptable range. You might be relying on the internal clamping diodes of the Arduino to clamp the logic level to ~5V. Since there are current limit resistors in place it the internal clamping diodes might be able to handle the higher input voltage all day, but is not optimal. Consider using a voltage divider which will output a lower nominal voltage.

If you have the datasheet for the illuminated switch you want to use, post a link to it and I'll update my answer to include how to wire it up.

Your diagram shows SPDT, so that's what you need. SPST isn't going to work.

The peak current through the switch will be the same as if you connected the relay coil directly to the 12V supply without the capacitor in place. (The capacitor merely causes the current to decay down to zero from this peak value.)

Therefore, if your 12V relay has a coil resistance of 400 Ω, the peak current will be 12V/400Ω = 0.030A.

With a 220 µF capacitor, the relay will activate for roughly 50-100 ms each time the toggle switch changes state. The exact value depends on the holding-current requirements of the relay.