Dave, I understand. I made some general comments as well, and I thought maybe an analysis of that design might be interesting to the OP and useful to others who might browse the thread later.
My main comment is that with a linear regulator there will be an issue with heat dissipation. Assuming say a 4V overhead for dropout, the input voltage needs to be ~16V. At the minimum output voltage, and maximum output current, the regulator will be dissipating 14.5V * 2.5A which is about 40W. To limit the case temperature to 100 degrees Celsius he will need a heatsink with a thermal resistance of about 2 degrees Celsius per watt to ambient, which is quite large.
My recommendation would be to buy an adjustable switching regulator module with up to 12V output and rated for 3A, and modify the feedback network to make it switchable and adjustable. These are available pretty cheaply I think.
Re the PC power supply, to the OP, trace out the secondary circuits. It will probably be possible to decouple the 5V and 12V circuits from each other and put them in series, to get a 17VDC output that can deliver 6A. That would be ideal to feed into the variable switching regulator I described.
Adding an adjustable current limit to the switching regulator wouldn't be easy; he would probably have to stick with the limiting built into the switching regulator IC. Other current limiting could be done with a series resistor (not a proper current limit, but I think he wants to use it for charging batteries, so it might be workable).
Regarding decoupling the 5V and 12V outputs. If you trace out the circuit of the secondary side of the PC power supply, you should find several independent circuits. One for 5V/18A, one for 12V/6A, and one (or maybe two) for the negative rails. These circuits will be tied together using a common ground rail and you'll have to isolate them from each other. There is a voltage detection and feedback circuit which is almost certainly connected across the 5V rail, so it's proably best to leave the 5V circuit as-is and isolate the 12V circuit. You really need to trace out the schematic of the whole secondary circuit, take a good photo of the tracks, and annotate that picture with the positions of the components in the secondary circuit. Then I, or someone else here, could work out how to isolate the 12V circuit and connect the outputs in series. This is not really a beginner's project but I'm sure it can be done, and would be a pretty interesting and rewarding project.
To the OP, re the two trimpots you found, the second one that doesn't seem to do anything will be the current limit trimpot. If possible, return it to its original position.
Oh, and be SURE to follow Steve's advice re safety! You may want to add a "bleeder" resistor across the main electrolytic on the primary side to make sure it discharges within a short time after power is removed. It will probably charge up to about 340V (mains voltage * 1.414 * 2 plus a bit). If you use, say, a 5W resistor and run it at 2W, the resistance is calculated as R = V^2 / P which is 340^2/2 which is about 56 kilohms. It will get hot while the power supply is running, so mount it away from the board, preferably with mechanical support. Assuming the input electrolytic is 200 uF (the value is not standardised but this is typical), the capacitor will discharge to a safe voltage after about two minutes from power being removed, and no screwdrivers need to be harmed. Another option would be to put a pushbutton in series with the bleeder resistor and only press it when you need to discharge the capacitor; that stops the bleeder resistor from getting hot during operation and you can use a lower value, e.g. 22K 5W which will bleed it quicker. In either case, use a multimeter across the electrolytic to check it's well discharged before touching the board.
