If the overhead is X, and the amount of work that can be done ignoring it is Y, you have Y - X. If you double the amount of work that can be down, but overhead stays the same you have 2Y - X, which is more than 2(Y - X) = 2Y - 2X. You're over performing doubling by not having to double the synchronization cost.

Not necessarily.

There can be overhead introduced between the "1 thread" and ">1 thread" cases.

From 1 -> 2 threads the overhead is introduced, if the cost is somewhat fixed, the gains from 2 -> 4 can be proportionally larger.