Here is a simple topology. This is opimized for simplicity at the expense of efficiency and effective use of the transformer's capacity:
With just a bare transformer secondary, you only get half-wave rectification to make each of the + and - rails. A center tapped secondary would allow full-wave rectification of both rails.
For a short time once per power line cycle, D2 conducts and charges up C2. Likewise, D1 conducts on the opposite polarity of the AC voltage and charges up C1. C2 and C1 hold up the rails between the bursts of current delivered thru the diodes. The two regulators then produce the nice and clean +12 and -12 volts. C3 and C4 are small ceramic caps that work better at high frequencies than the larger electrolytic C2 and C1. These help keep the regulators stable. C5 and C6 are generally required by such regulators, and again help keep them stable.
The only real brain cycles required here is to calculate what C2 and C1 need to be. At 60 Hz, each will be charged up every 16.7 ms. For sake of analisys, let's simplify this to assume C2 is charged up instantly to the full voltage once every 16.7 ms with nothing being added to in at other times. That's actually a little worse than what really happens, so is a good thing to design to.
You say the maximum output current is 500 mA. A 15 VAC transformer will put out 21.2 V peak. Let's be pessimistic and say the diode drops 1 V at the full current. That leaves 20.2 V that C2 gets charged up to. 78xx regulators need a lot of headroom, let's say 2.5 V (your job to check the datasheet). That means the voltage on C2 can't be allowed to drop below 14.5 V. 20.2 V - 14.5 V = 5.7 V that C2 is allowed to drop in the 16.7 ms between getting recharged. (500 mA)(16.7 ms)/(5.7 V) = 1.5 mF. That should be at least a 30 V cap. The same logic holds for C1.
This should all work, but you also need to look at the power dissipation. This simple circuit won't be very efficient. From above, we can see that the average voltage on C2 at full current is 17.4 V, which means each regulator is dropping 5.4 V. That times the 500 mA current is 2.7 W each. A TO-220 package can handle that, but will require at least a small heat sink.