Why is the lower bound $m \log n$ for this make-set, union and find-set sequence?
You are asking two questions.


  Is the lower bound only for this specific implementation?


Yes. If you also use path compression, the running time will be $o(m\log n)$.


  The lower bound seems to large to me.


Your mistake is that you assume that there are only $\log n$ UNION operations, whereas there are $2^{\lfloor \log_2 n \rfloor}-1 = \Theta(n)$, as the answer clearly .