I honestly think that you need to check in-house with whoever understands this procedure, because as you present it, it does not make a lot of sense.
If the standard has 0.2 % IPA, presumably the samples do as well, so the acetone is 99.8 % if there are no other impurities. Unless you do multiple injections and take means, a GC split injection is not repeatable enough to distinguish between 99.8 % and 100 %. Therefore, the area of the acetone peak cannot be used directly to determine acetone concentration.
If the response factor corrects for the difference in response of the FID to the two different substances, then with a single standard you can just as well take the IPA: acetone peak area ratio for the sample, divide it by the ratio for the standard and multiply by 0.2% to get the IPA concentration in the same units as in the standard (V/V).
You could take acetone concentration as 100% - IPA%, but as you have noticed this will only work if there are no other impurities. A major problem with 100% - IPA is that the FID does not detect water, which there is a good chance that the acetone contains.
Peter