Do the minimum spanning trees of a graph have the same number of edges with a given weight?
The question and this post refers to this answer at its version 2.


  First of all, what does minimum disagreeing weight mean?


The better word should have been "distinguishing". If the set of edges of some fixed weight in $T_1$ is different from that of $T_2$, then $T_1$ and $T_2$ disagree on that weight, i.e., that weight is distinguishing the two minimal trees. 


  What is supposed to be $g(e)$?


That is a typo. It should have been $w(e)$.


  Why is $w(e') \lt w(e)$ here? isn't $e'$ same as $e$ ? and why does  $T_1,T_2$ contradict as minimal spanning trees?


Let us use a new dummy variable instead of $e'$ for the last part of that answer.

Let $f \in C_{T_1}(e) \cap T_2$. Had $f$ weighed the same as $e$, we should have update $T_1$ by using $f$ instead of $e$. Since we have done all such updates, $f$ must weigh differently from $e$. That is the conclusion in the previous paragraph, "all edges in $T_2 \cap C_{T_1}(e)$ (where $T_1$ is the updated version) have weights other than $w(e)$".

There are two disjoint cases.


the case when $w(f)\lt w(e)$. Consider the spanning tree $(T_1 \setminus \{e\}) \cup \{f\}$, which is lighter than $T_1$.
the case when $w(f)\gt w(e)$. Consider the spanning tree $(T_2 \setminus \{f\}) \cup \{e\}$, which is lighter than $T_2$.


In either case, we have found a spanning tree that is lighter than a minimum spanning tree, which is a contradiction.