I know this can be done for DC, but if i choose AC (undecided) load, can a resistor be added to AC too? i thought resistors handled only minimal current.
It will work in the same way for AC. As long as you work only with resistive elements (heaters, lamps, resistors), not capacitors or inductive coils, any DC calculations can directly be transferred to AC, using the "effective" (root mean square) values for AC voltage and current.
AC is not automatically higher current than DC. But yes, if you want to power a heater and run it on a low voltage, you will probably deal with currents of several Ampere, and need to size your resistors appropriately (see below). The standard resistors used in small-signal circuits will indeed be specified for somewhere between 1/8W and 0.6W only, but you easily can get resistors specified at 10W (and much more, but eventually they start to get expensive). See e.g.
https://www.reichelt.de/11-Watt-axial/2/index.html?ACTION=2&GROUPID=3120https://www.reichelt.de/from-50-Watt-axial/2/index.html?ACTION=2&LA=2&GROUPID=5274(Switch to English in the upper left, if the site doesn'a already come up that way.)
By the way, if you already have the current sensor modules for DC, I would still suggest powering the heaters by DC for simplicity. If you need 2*25W of heating power and use 12V heaters, that's less than 5A total. You can get 12V 5A power supplies cheaply, for use with LED strips or halogen spots. Higher powers, e.g. 12V 10A, are also available. The Adafruit current sensor will limit your power, due to its 3.2A limit.
I just need to calculate the resistor value. i will probably run at 12V or 5V depending on my available power supplies. V=IR or P=V2/R? i have never sized a resistor before. i suppose this will need to be done later after i find out the difference.
Yes, that's how it works.
You mentioned earlier that you want at least 25W heating power, and your Adafruit current sensor can only handle 3.2A. So you will need to run at 12V, since 5V * 3.2A will not give you enough power.
Assuming a heater module specified at 12V and 30W, that would run at 2.5A, and would hence have a resistance of R = U/I = 4.8 Ohm. Assuming that the variation between units is no more than 20%, your matching resistor will be anywhere between 0 Ohm and 1 Ohm. For the power rating, the highest resistance is the worst case. P = I² R = (2.5A)² * 1 Ohm = 6.25W.
I would use a 15W to 25W resistor in this case, because it will run hot inside its insulated box. If the resistance you need to match your two heaters/lamps is smaller, the required power rating will also be lower.