A home work question! But you declare it as such, and you already understand the concept - in this scenario (a voltage source) current isn't "pushed", it's "pulled" by the load. So the short answer here is that you'll be OK buying the 25 V, 2 A transformer.
A different way of thinking about this is for both voltage and current, the listed rating is what is/can be supplied. Therefore a 25 V, 2 A transformer will have 25 VAC on the output and the maximum current which can be drawn is 2 A. If the load draws less than 2 A, the no problems. More current is beyond what can be supplied, which will mean the transformer will overheat, with the follow-on consequences you might imagine.
You can't use the 12 V transformer because - assuming you're using a linear regulator - you need to supply a higher voltage to the regulator than what it regulates to. Exactly how much higher is a question for reviewing the data sheet for the device.
The LM358 OpAmp is current limited to typically 40mA.
With a +-5V supply you are limited to +3.5V and -5V output excursion.
If you short the output you will source 3.5/50=70mA or sink -5/50=-100mA into the grounded 50Ohm series resistance. This will result in a max of 0.245W (when positive) or 0.5W (when negative) dissipation.
Unless you have a constant -5V output that is shorted you will see an average that is less than 0.5W dissipated, if you remain in the linear portion of the output (+3.5V -3.5V) you will have less than 0.25W into a short. You do not need extra short circuit protection at those supply voltages.
The OpAmp is rated continuous short circuit proof with less than 15V supply.
If you have it calibrated for 2V open circuit you will see 1V across a 50 Ohm load and have a matched source impedance and not be able to approach any device limits.
I would suggest fast diodes connected from the output to the supply rails to prevent external devices from causing device limits to be exceeded.
EDIT:
The max current limits calculated cannot be achieved with this particular OpAmp as it lists a short circuit output current of 40mA typical 60mA max and safe for continuous short circuit, it is inherently protected and the current limit protects the output resistor. Higher output currents could be reached with some other types. The max output voltage is listed as the positive supply - 1.5V hence the 3.5V positive limit with a +5V supply, devices that can swing closer to the supply rails are also available and have their uses.
All of the numbers used are available in the data sheet, either in the text, tables or graphs. Note 1 on Table 1 warns about short circuit dissipation limits with supplies over 15V
Calibration at 2V was just my suggestion for the choice of gain components so full-scale digital output would be calculated to give 2V output or 1V into a matched 50 Ohm load, these low voltages would also be self protected as the currents would be even less and within the 30mA supply limits to maintain accurate operation.
I keep forgetting to point out that the possible supply current limits of 30mA would be reached before the resistor dissipation or OpAmp limits. This may cause unexpected behaviour especially if one supply rail were to be reduced more than the other for instance, though with this device this is less likely as it is a tracking dual regulator. It is possible to configure the Mitsubishi M5290P for more than 30mA with external transistors so it is not certain that is will be current limiting at 30mA in this re-purposed power supply circuit.
Best Answer
If the simulator model for the U2 amplifier doesn't include a current limit, the intended effect of Q2 is defeated. Put a series resistor in the output, between the C1 and collector connections, to get a real current-limit effect.
Simulators depict real devices, but aren't (internally) as complicated as reality. As others have pointed out, 2N2222 won't carry the intended current.