Short: Add a 1 ohm resistor in series with the transformer :-).
Longer:
A "perfect" transformer and 'perfect" capacitor will have infinite current spikes, as I know you realise.
While real world results will vary with transformer maker's 'ethos and philosophy', the real world experience is that you wil usually get superior results by adding a small "conduction angle spreading resistor" in series with the transformer winding feed to the capacitors. This is counter intuitive to what you may expect from an efficiency point of view and is often not done in practice. Theoretical calculation of the effect of such a resistor is surprisingly annoying but simulation will show the effects instantly.
Given that the mean DC level under load is 0.7071 ( = sqrt(2) ) of V peak, you have quite a lot of headroom to work with and can afford a modest amount of drop in the series resistance. There are several scondary effects which may be useful depending on environment. Spreading the conduction angle improves the power factor of the otherwise very peaked load - but probably not enough to make a difference in meeting or failing formal power factor requirements. Sometimes more importantly, spreading the conduction angle greatly reduces peak loads on the diodes and reduces EMC issues (ie less radiated electromagnetic noise) - probably not an intuitive effect of adding a few ohms of series resistance.
Lets have a play with some figures:
You have 15 VAC secondary voltage and are aiming at 12VDC at 2A.
Assume for now that about 15VDC minimum on the filter caps is acceptable 9giving the regulator 3V headroom minimum).
Vpeak is 15 x 1.414 = 21.2 V
Load power is VI = 12 x 2 = 24 Watts.
If you managed to filter this well enough to achieve say about 20VDC on the cap you would dissipate Vdrop x I = (20-12)x 2 = 16 Watts in the regulator and "as a bonus" achieve massive ripple CURRENT in the caps but little ripple VOLTAGE. This does not seem like a marvellous idea :-).
If you can manage to spread conduction over 25% of the voltage cycle you will get mean current during conduction down to 4 x Iavg = 8A.
Assuming 21V peak, 25% conduction occurs at about 19V transformer output, and a very useful 50% conduction happens at just under 15V. See graph below.
This suggests that inserting even one ohm series resistance is going to have a substantial effect. If the 8A mean that is required for 25% conduction is dropped across 1 ohm the 8 volt voltage drop is going to ensure that the 8A does not happen (as 21-8 = 13V which is lower than the 15V DC target this was based on).
If 50% conduction occurs then mean current during this period will be 4A and mean drop across 1 ohm would be 4V so this may be "about right" as if the filter cap was at about 15V you'd get (21-15)/1 = 6A peak at waveform peak - and as the cap will have "rippled up" in voltage by then you'll get less than 6A). And so on.
Yes, you can analytically work out what happens. But, just put 1 ohm in the simulator and see what happens.
This has the effect of putting MORE ripple voltage on the capacitor(s), LESS ripple current, less regulator losses and less transformer losses, less diode EMI.
The series resistnce could be in the transformer but then addes to heat generatoion inside a relatively costly component where you'd rather be trying to optimise power transfer rather than heat loss. A 5 Watt 1 ohm resistor will probably work OK here. 10W would be safer due to peaks. eg 4A at 50% = I^2R x 50% = 15=6W x 0.4 = 8W BUT waveform is complex so actual heating needs to be calculated.
Note that in many cases the ripple current rating of two capacitors is superior to that of a single capacitor of equal total capacitance.
Use 105C (or better) caps as a matter of course in this sort of application. 2000 hours+ a good idea. Cap life ~~~ 2^((Trated-tactual)/10) x Rated_life
Switching an inductor may make it work like a boost converter, and thus temporarily generate a higher voltage.
One way of making sure that the input voltage to the LM338 converter is no higher than allowed, is to wire a high-wattage (between 600W and 5 kW) TVS diode with an appropriate clamping max voltage biased against the input voltage and ground.
Also, if you're switching on the measured/actual output voltage, then you can get into oscillation. It would be safer to switch on "desired" output voltage, and let the regulator make sure it gets there. You could perhaps feed the bias voltage from the regulator ADJ into your microcontroller with a resistive divider to achieve this. Or you could just use a comparator with a pre-set trip point to drive/un-drive the relay.
Best Answer
You want to go with the transformer with the 18VACRMS rated secondary.
A transformer with that rating will have a peak output of 18/0.7071 = 25.46V peak.
The rectifier diode will contribute about 1V of voltage drop to the 1000uF filter capacitors. So I would expect the filter caps to see a peak value of ~24.5V.
During operation as the load is drawing power from the filter caps the voltage will discharge the caps by some amount less than the 24.5V peak. It is a very good guess that the average value of the voltage at the filter caps is the 22V as shown on the schematic.
The beefy transistors that are inline with the current paths to the outputs are the linear regulator pass transistors. The TA7179 chip regulates the base current of these pass transistors to achieve the +/-15V outputs through variations in the load current.
I built a simulation of the transformer secondary and the diodes plus filter capacitors. Then I added a load resistor to each of the outputs. Here is the simulation circuit:
Note that the voltage sources have been specified at a sine wave peak of 25.46V. The two sources have been arranged to be 180 degrees out of phase as you would expect for a center tapped transformer. The load resistors were selected at 22 ohms to represent a nominal 1A load at 22 volts.
Here you can see the positive voltage at the filter capacitor and the current load into 22 ohms. Note the average value of the voltage is almost on 22V as labeled on your schematic.
In like manner this shows the corresponding negative filter capacitor voltage and the current load.