You are right. There is more than one way to prove this, one of them is with a potential function. This link (and many others) explain the potential method. However, textbooks usually require that the initial value of the potential function is 0. Let's generalise for the case that it is not.
For the binary counter, the potential function of the counter is the number of bits set to 1. When you increment, you spend k+1 time to flip k 1's to 0 and one 0 to 1. The potential decreases by k-1. So the amortised time of this increment = ActualTime+(PotentialAfter-PotentialBefore) = k+1-(k-1) = 2 (constant).
Now look at the section "Relation between amortized and actual time" in the wikipedia link.
TotalAmortizedTime = TotalActualTime + SumOfChangesToPotential
Since the SumOfChangesToPotential is telescoping, it is equal to FinalPotential-InitialPotential. So:
TotalAmortizedTime = TotalActualTime + FinalPotential-InitialPotential
Which gives:
TotalActualTime = TotalAmortizedTime - FinalPotential + InitialPotential <= TotalAmortizedTime + InitialPotential
So, as you say, the total time for a sequence of k increments starting with n is O(log n + k).