> Insertion sort is O(n^2) on the number of comparison operations, because this step is O(n) on the number of sorted items

Locating the correct position in the sorted items is O(log n) on the number of sorted items. You point this out yourself later in your comment. Because the sorted items are sorted, it's not necessary to examine each of them.

Doing O(log i) work as i varies from 1 to n is O(log n!) work, which is O(n log n); not much different from doing O(log n) work as i varies from 1 to n.

I'm not sure how to interpret the two halves of this quote. It looks to me like I'm doing O(n log n) comparisons, and also finishing the sort in O(n log n) work overall. I'm not taking O(n^2 log n log n) work to finish the sort.