The answer is different for a variable wattage device vs. a mech mod.
For a variable wattage device, power in is equal to power out, with some loss for efficiency of the circuitry. If you are putting out 30W, and all of that power is being drawn from a single battery, then that battery must supply 30W of power, plus some overhead. Let's call it 36W figuring a safe 20% efficiency loss. A mostly discharged battery will put out around 3.7V. For a regulated device, which is always trying to draw a fixed wattage, you should calculate A based on a discharged battery, because A will be highest when V is lowest for a fixed wattage. Since W = I * V, we have I = W / V, or I = 36 / 3.7, which is just under 10A. This is right on the limit of what is safe. I think it's probably OK, but I would prefer to have a 15A or 20A battery, just to be on the safe side. If you were trying to put out, say, 50W, then that would be 60W with efficiency loss, and you'd need 60/3.7, which is a little over 16A. So, you'd still be safe with a 20A battery, but would be over-discharging a 10A or 15A battery, which is unsafe.
On the other hand, if you're only drawing 20W, call it 24W with overhead, you'd need 24W/3.7V which is 6.5A. Then, a 10A battery would be perfectly safe.
If your variable wattage device is powered by two batteries instead of one battery, the power is drawn equally from both batteries. So, halve the above A numbers. All of the above is pretty much true for the whole range of resistances, though you might not be able to actually set the device to higher wattages if you have a resistance that's too high. This will vary from regulated device to regulated device.
For a mech mod, the battery will always try to supply between 3.7V and 4.2V (3.7V when it's drained, and 4.2V when it's fully charged). The amps that are drawn from the battery will be inversely proportional to the resistance, since I = V / R. Calculate all I and W values figuring a fully charged battery, because that's where A will be highest. For a 1 Ohm coil, and a battery producing 4.2V, I is 4.2/1 which is 4.2A (4.2V * 4.2A is a little under 18W). For a 1.2 Ohm coil, 4.2/1.2 is only 3.5A, and will produce a little under 15W. If you go all the way down to 0.5 Ohms (not recommended for a Kayfun), you'd be pulling 4.2V/0.5 Ohms which is 8.4A, and is a little over 35W.
So, my recommendation would be to get at least 20A or better continuous discharge batteries. For lower wattages, you're probably safe with a 10A battery, but what if you decide you want more wattage? You'll probably forget that your battery can't handle it, and end up over-discharging your battery. Baditude has a blog that lists safe ratings for a variety of batteries:
Baditude Battery Basics Blog
Pretty much any of the batteries he has in bold are good choices. Be sure to select 18650 batteries if that's what your device needs.
Hope that helps!
Found who wrote the saw manual!
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