It's almost correct. To have a better approximation you should calculate with the nominal battery voltage, which is 3.7V for our cells. So "full cells" means you have ca. 7.4V under load (until the battery gets used a while, then less). As soon as you hit the fire button your 4.2V
batteries won't have 4.2V any more.
So the fully charged calculation should be with 148W, so on the firing side with 90% efficiency you'd have 133.2W left as max. power if you don't want to draw more that 20A from your
batteries.
Now let's check what happens at the lowest voltage, let's say 3.2V cutoff again (on battery side):
148W/2=74W per battery
74W/3.2V = 23.125A <- maybe a problem for your component, you will need to regulate power down
That component you linked looks good, but it can only go up to 110W. Now we're almost at the 100W I suggested in my 1st post
Also keep in mind that your
batteries will need to power your atmega + your voltmeter (+ your Ω-meter) and lose heat there as well, so the efficiency will go down further.