You are incorrectMosher's proof is, I think, fine. Re You should re-read the definition of $i(\delta, \{h\})$ given on line -14 of page 38 of Mosher's paper.
The sequence of flips produced by Mosher's algorithm is at most linear in the total intersection number. This worst-case bound is realized when the two triangulations differ by a power of a Dehn twist. (An average-case case bound is also possible; here the number of flips will be the log of the total intersection number.)
Another algorithm can be deduced from Hatcher's paper "Triangulations of surfaces".
In reply to the updated question:
Let $d_1$ be the first arc $h$ crosses, and let $d_2$ be the second. So $d_1$ is the diagonal of $Q$, and $d_1$ will be flipped to $d_1'$. After the flip, the first arc that $h$ meets is $d_2$. So the intersections of $h$ and $d_2$ are not counted. Thus $i(\delta',\{h\}) \leq i(\delta,\{h\}) - 1$, as desired.