Your calculations are correct. A 120 Ohm, 2W resistor is fine for the 12V rail. A 4W resistor might not get as hot if it is physically larger.
The spec sheet implies a minimum current for each of the 3 outputs, but you may get reliable operation with less load than this implies. For the 5A rail, you would need a 10 ohm, 10 watt load to keep it in spec, but the spec also relates to the regulation (+/- 1%) and you probably don't need such a load to keep the PSU turned on.
You need to apply a dummy load to account for the worst-case operating conditions, but can safely use a higher current load temporarily. If you want to experiment and save some power, I'd try ~220 Ohm on each rail, and see if that made it happy (but I'm guessing here).
You're on the right track in two regards; your calculations headed in the right direction, and the Radio Shack guy doesn't know what he's talking about. A general good start in thinking about electrics is to consider vague magnitude. You have a project dealing with 40W here. Does 1/8 of a Watt sound like the right sort of answer?
The bit of calculation you need here is the power dissipation in the resistor. A series resistance circut (when resistors are connected end to end, "in series") is pretty simple. In this case, we know that the current flowing in the circuit is 0.335A or so. We also know that the resistor you are using is 100 Ohms.
The power dissipation in a component is Volts X Amps. Combining this with Ohms Law, we can show that this power is also Amps squared X the resistance in Ohms.
This gives us 0.335 X 0.335 X 100. Which gives an answer of 10.77 Watts.
So the vague magnitude speculation actually gave you a pretty good answer, calculating it we find a 20W rated resistor will comfortably handle this load. And not to trust the guy in Radio Shack.
edit Oh poo, I didn't realise you have two parallel strings. And now I'm typing this retraction I can't see your picture. BRB.
edit 2.
Okay, we now have 50 lamps across 120V. The total resistance of all the lamps is V*V/P, which is 120*120/20, or 120*6, which is 720 Ohms. Divide that by 50 to get the resistance of 1 lamp, which is 14.4 Ohms.
Now the resistor needs to be the same resistance as 27 lamps, so it's 27*14.4 which is about 388.8 Ohms. So that's our resistor.
The current flowing is I/R, which is 120/720, about 0.166 Amps. And the power dissipated by the resistor is 0.166*0.166*388. Which comes back to 10.77 Watts again; which it ought to because we're dissipating about (rule of thumb) 1/4 of the entire string's rated power as useless heat in the resistor.
Best Answer
I've used a 10 Ohm 5 Watt resistor in the past and it works well for testing both Zinc-Carbon and Alkaline cells - both "AA" and "AAA".
You will have a current up to 160 mA which is enough to load the cell and show its true voltage at that current level.
A new cell will have a voltage exceeding 1.5 Vdc (often above 1.6 Vdc); a used but still usable cell will have a voltage between 1.2 to 1.5 Vdc; a weak cell will have a voltage between 1.0 to 1.2 Vdc.
A cell with a loaded voltage less than 1.0 Vdc is considered to be dead. Some stuff might run from it but not for long.