There are some real engineers here that can calculate amount of capacitance needed for amount of ripple at current, but you didn't mention the current your circuit needs to be capable of. However with 7812, some would use a generic rule like 1000uF between transformer and that regulator for each amp current.
The cap after it, could be anything low ESR like a 10uF tantalum, but if there is a long run to the powered device, it should have capacitor(s) local to it too.
You're starting with a quite high 25V input. Put that through a bridge rectifier and you end up with around 1.4X that minus two diode forward drop losses, so around 34V, regulated down to 12V, is a SUBSTANTIAL loss. For this reason, I suggest that your PSU circuit proposed is not very good for more than a few tens of mA current and really, long term it's still better to find a lower voltage transformer.
Ideally you only want your transformer rated for a couple volts (technically more like 2.5V IIRC) over your output regulated voltage at max forward drop for the regulator or instead you can calculate based on max actual load it will see (see its datasheet, should have a graph about Vf vs current). This will reduce power loss, meaning less wasted electricity on your power bill, and less heat to deal with which linearly reduces the amount of heatsinking needed. Voltage drop * current = watts heat dissipation. So, with a 1.2A load, at a 22V drop, that's 26W lost as heat, way too much to be reasonable.
Must you use this center tap 25V output? If so, then I would insert a buck switching regulator between the bridge rectifier and the linear regulator to more efficiently lose most of that excess voltage, if you need cleaner power than a SMPS output produces (so can't just get rid of the linear 7812 regulator and use straight from the buck SMPS output) and can't calculate a sufficient filter to quiet down the ripple.