Ci and Co in fig 7 here
https://www.google.com/url?sa=t&rct...CmnUPkZ9qd8A&bvm=bv.72197243,d.aWw&cad=rjaare decoupling and MUST be mounted close to the LM317T. So, C1 - yes. Confusing drawn schematic like Nigel said.
The DC voltage will be about 33.6V (24 * 1.4) if it's indeed 24V. It's close the absolute max input of 40 V. Surge suppression is in order for a reliable circuit.
The regulator will drop about (33-12 V) * 0.6A or 12 W. The regulator might be getting too hot and shutting down. It will limit start-up currents to 1.5 A.
The temperature and actual output should be looked at under load. You have to have a minimum of 3V across the regulator for the regulator to work.
Here
https://www.google.com/url?sa=t&rct...FEfkADk5oUMa3KwjgqL3-YA&bvm=bv.72197243,d.aWw your looking at full wave-capacitor input, and you do get a derating of the current x 0.62. 40/24*.062 is about an amp, so you should be OK.
I'd like to know what part number you used for the bridge and which you used for the alarm.
The most likely cause is that the regulator is shutting down thermally, not heat sinked properly and definately the secondary voltage is too high.
Vsec should be > (13.9+1.2+3)/1.4 or about 12.96 VAC. Those numbers come from sqrt(2) which takes and RMS voltage and gets a 0-peak value. The 13.9 V is what you want to regulate at. The 1.2 is the voltage drop of 2 diodes in the bridge and 3 V is required for regulation. As you can see, your transformer is about 2X the voltage required.
The circuit should be happy with a 40 VA 12 or 12.6 V transformer.
You could try feeding it with DC from a laptop power brick which is about 19V or so and see if the circuit behaves. You can feed the DC into the bridge rectifier.
There are other losses from rds(ON), but they should be ignoreable.
So, I'll bet the regulator is getting too hot.