No, Kruskal's algorithm doesn't make any assumptions about how fast edges can be sorted. I suspect you misinterpreted what you read. Kruskal's algorithm doesn't make assumptions.
You linked to a Wikipedia article. It states:
Provided that the edges are either already sorted or can be sorted in linear time (for example with counting sort or radix sort), the algorithm can use a more sophisticated disjoint-set data structure to run in O(E α(V)) time
"Provided" means "If", not "We assume that".
If you use a standard sorting algorithm that takes $O(E \log E)$ time, then Kruskal's algorithm runs in $O(E \log E)$ time. If you are in a special situation where there is a way to sort the edges in linear time (e.g., using counting sort), then Kruskal's algorithm can be implemented in a way that takes $O(E \alpha(V))$ time, which is faster. But there is no assumption that you will always be in that special situation; usually, you won't.
The case where there are only two edge weights is indeed such a special situation, and thus Kruskal's algorithm can be implemented in a way that is asymptotically faster for such graphs than for general graphs.