There is hidden linear time to perform merge part.
Linear is less than loglinear, so it was not visible. $\log1=0$, but it means that you will not e.g. heapify, just merge in $O(n)$.
Also all other complexities hide constant factors, so $O(\log k + c)$ drops constant part, but when it becomes 0, you should reveal hidden part and say it is $O(1)$.

There is hidden linear time to perform merge part of sorting algorithm.
Linear is less than loglinear, so it was not visible. $\log1=0$, but it means that you will not e.g. heapify, just merge in $O(n)$ (heap structure in case of 1-sorted array degrades to swapping consecutive elements to perform sort).

Inserting element to empty heap (which might be used in such sorting) costs $O(1)$, while putting element to min-heap containing $k$ elements is $O(\log k)$, which is in fact $O(\log k + c)$, so the first looks like it could be substituted with $k=1$ to have cost 0, but the second reveals constant factor in asymptotic notation.

There is hidden linear time to perform merge part.
Linear is less than loglinear, so it was not visible. $log1$ is equal 0, but it means that you will not e.g. heapify, just merge in O(n).
Also all other complexities hide constant factors, so $O(logk + c)$ drops constant part, but when it becomes 0, you should reveal hidden part and say it is $O(1)$.

There is hidden linear time to perform merge part.
Linear is less than loglinear, so it was not visible. $\log1=0$, but it means that you will not e.g. heapify, just merge in $O(n)$.
Also all other complexities hide constant factors, so $O(\log k + c)$ drops constant part, but when it becomes 0, you should reveal hidden part and say it is $O(1)$.