Hi Steven,
My wifey used to nag me, "Are you making another one?", "Your LEDs are blinding me" and "Get off the internet". But now that she has taken up knitting and going to cooking school, she doesn't bother me very much!
Ah, now we get to know most of the details of your circuit, except the most important one: What is the voltage across the non-functioning relay coil? But now we can calculate it from the relay coil's resistance and the 555's and 1N4148's spec's.
1) Your battery was dead. I suspected it but even a good battery doesn't help.
2) The 40 ohm relay coil is like a dead short to a little 555 and a little 1N4148. The resistance of the coils in the 12V relays that I use measure 1K and 2.2K ohms.
3) The 555 is rated for up to 200mA of output current. With 200mA, its output typically has a 2.5V loss. Your 40 ohm relay will draw 300mA from 12V. Therefore the 555 will probably have a 3V loss or more. A 3V loss at 300mA results in 900mW of dissipation, which is very close to the 555's maximum temperature rating. It might melt.
4) The 1N4148 is rated for up to 100mA continuous current. With the relay's 300mA it will typically have a voltage drop of 1.1V or more, and might also melt.
So the circuit's voltage loss is 3V or more for the 555 and 1.1V or more for the 1N4148. If you subtract those losses from your 12.58V battery, you end up with only 8.48V or less across the relay coil.
A 12V relay is guaranteed to work with 9V or more.
That's why your circuit doesn't work. Ohm's Law, a couple of measurements and the datasheets explain it all.
When you use a relay with such a low coil resistance, you're going to need a power transistor to drive it.