Double coin toss. We wanna know what's the probability of throwing at least one head.
First we model it as $\Omega$ = {{H $\cap$ H}, {H $\cap$ T}, {T $\cap$ H}, {T $\cap$ T}}. I often see here that people solve: P({H $\cap$ H} $\cup$ {H $\cap$ T} $\cup$ {T $\cap$ H}) = P(H $\cap$ H) + P(H $\cap$ T) + P(T $\cap$ H). But this is not possible because Kolmogorov's axiom #3 just allows addition of disjunct events/sets, but (H $\cap$ T) and (T $\cap$ H) are not disjunct. So in this model we'd need a different way to answer the question, e.g. thru using the probability of the complement.
Now we model it as $\Omega$ = {{H1 $\cap$ H2}, {H1 $\cap$ T2}, {T1 $\cap$ H2}, {T1 $\cap$ T2}}. Now we can apply the addition rule: P({H1 $\cap$ H2} $\cup$ {H1 $\cap$ T2} $\cup$ {T1 $\cap$ H2}) = P(H1 $\cap$ H2) + P(H1 $\cap$ T2) + P(T1 $\cap$ H2) because all events/sets are disjunctive.
In model 1, Bayes' theorem holds, because (H $\cap$ T) = (T $\cap$ H) by commutativity and therefore its probabilites are the same which is the condition to derive Bayes' theorem. But in model 2, Bayes' theorem does not hold because you do not have a relation of commutativity like in 1, but you need it, else it's not a theorem there.
Can you tell me if I am right in all three points, and if not why?
If a student wrote something like (1), I'd assume they meant (2) but weren't being careful with notation. I'd encourage them to define the events in words before doing any symbolic reasoning. ("Let $H_1$ be the event that the first toss comes up heads," etc.)
For (2), assuming the definitions of $H_1,H_2,T_1,T_2$ as above, you're right that the union is disjoint. We can treat this as obvious or prove it from the fact that $H_1,T_1$ are disjoint and $H_2,T_2$ are disjoint.
As for (3), Bayes' theorem holds for any pair of events. Of course, $H_1$ and $H_2$ are not the same event, so it would be a mistake to treat them as such (by accidentally calling them both $H$, for example) when applying the theorem.