Disjoint union topology vs Product topology

Not quite. First a minor point that may or may not help to solve some of your confusion. If $a$ is a subset of $A$ and $b$ is a subset of $B$, then it does not make sense to write $a \oplus b \in A \oplus B$, but $a \oplus b \subseteq A \oplus B$ would make sense; $a \oplus b$ is not an element of $A \oplus B$ but a subset thereof. Similarly, writing $i_A(a) = a \oplus \emptyset$ is problematic, since $a \oplus \emptyset$ is not an element of $A \oplus B$. Moreover, it need not be the case that one of $a$ or $b$ is necessarily empty. For instance, if $A = \{1,2,3\}$ and $B = \{4,5\}$, then $\{1, 5\} \subseteq A \oplus B$ but both $i_A^{-1}(\{1,5\}) = \{1\}$ and $i_B^{-1}(\{1,5\}) = \{5\}$ are non-empty.

Note that this really has nothing to do with the topologies; here we are only thinking about what $A \oplus B$ is like as a set. Indeed, figuring out the difference between $A \oplus B$ and $A \times B$ as sets would be a good starting point: if they are different sets, then certainly they will also be different topological spaces.

Take $A = \{1, 2, 3\}$ and $B = \{4, 5\}$ as above. Then $A \oplus B = \{1, 2, 3, 4, 5\}$ consists of $5$ points (but depending on your treatment of disjoint unions, you may want to write this as $A \oplus B = \{(1,0), (2,0), (3,0), (4,1), (5,1)\}$ or something like that). In any case, the set will consist of $5$ elements.)

On the other hand, $A \times B = \{(1,4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}$ consists of $6$ points so the two sets are really different (in that they have different cardinalities and there can be no bijection (thus in particular no homeomorphism) between them).

However, coming to the actual topologies, I can see where your confusion might arise. In both the case of the disjoint union topology and the product topology we have very natural maps: in the first case we have the inclusions, and in the second case we have the projections. When we cook up the two topologies the way we do, it ensures that in the first case, the inclusions become continuous for free, and in the second case, the projections become continuous for free, so certainly there are analogues between the two notions. You already seem to be aware of this though, as you mention products vs. coproducts.