Merge two series of sorted number, one much longer than the other
What exactly does $n \gg m$ mean for your professor?

If it means $m \in o(n)$, then you are allowed $O(\log m)$ time for step 2.2. Note that $\log(n/m) = \log n - \log m$. If $m \in o(n)$, the target runtime bound simplifies to $O(m \log n)$; a summand $m \log m$ is dominated by this.

If $m \in \Theta(n)$ is allowed -- arguably, $m = n \cdot 10^{-10}$ would fulfill "a lot smaller" -- you can have that  $\log n/m \in O(1)$ and the runtime bound is $O(m)$. It's quite clearly impossible to perform this task in time $O(m)$, so I think we can safely ignore this case for the purpose of this exercise.

But here is something more for you to think about.

In step 1, you need to specify "special"; if you don't pick the elements according to your sketch, the rest won't work out. So we have to select the elements $i(n/m)$. In order to do this in $o(n)$ time and perform binary search efficiently, we need the series to be stored as arrays.

We can not copy all values to a new array in the required time. We can not insert values into an array, either. So how is this to work at all?