Yes, you are correct. The topology is the final topology wrt the standard injections $\sigma_i, i \in I$ on the space $\coprod_{i \in I} X_i$ (defined as the standard set disjoint union construction).
It's easy to observe that each separate map $\sigma_{i_0}$ is also open, as $(\sigma_i)^{-1}[\sigma_{i_0}[O]]= O$ for $i=i_0$ and $\emptyset$ if $i \neq i_0$, and when $O \subseteq X_{i_0}$ is open, this set is open for all $i$ for and hence is sum-open in the final topology. So $\sigma_i[X_i]$ is homeomorphic to $X_i$ for all $i$, hence the identification.
And then you can reformulate $O$ being open in $\coprod_{i \in I} X_i$ as
$$\forall i \in I: O \cap \sigma_i[X_i] \text{ open in } \sigma[X_i]$$
because $$O \cap \sigma_i[X_i] = \sigma_i^{-1}[O]$$
So that $\coprod_{i \in I} X_i$ has the so-called coherent topology wrt its subspaces $\sigma_i[X_i]$ (after we've declared $\sigma_i$ to be a homeomorphism, so making the identification).
So this gives two views on the topology, that come down to the same idea in the end.
I have never seen "pairwise disjoint family indexed by $I$" used to mean Definition 1; it always means Definition 2 in my experience. Note though that you can also talk about a pairwise disjoint family where "family" just means "set" rather than "indexed set". In that case, Definition 1 is what it means for the set (not the indexed set) $\{A_i\}_{i\in I}$ to be pairwise disjoint. Incidentally, I strongly recommend not using the same notation for a set and an indexed set; I would use $(A_i)_{i\in I}$ for the indexed set (which is really a function with domain $I$) and $\{A_i\}_{i\in I}$ for the set (which is the image of that function). On a related note, if you are using "family" to mean indexed set rather than just set, your statement of the axiom of choice should be
If $(A_i)_{i\in I}$ is a family of pairwise disjoint, non-empty sets then there is $C$ such that for all $i\in I$, $|A_i\cap C|=1$.
The version you stated cannot be used with either of your definitions since neither definition defines what it means for a set $A$ (rather than an indexed set $(A_i)_{i\in I}$) to be a pairwise disjoint family. (Of course, you can make such a definition, and get a third version of the axiom of choice. It is trivial to see that this third version is equivalent to the axiom of choice using Definition 2, since given any non-indexed family $A$ you can make it an indexed family by the identity map $A\to A$ (that is, you take $I=A$ and $A_i=i$).)
In any case, the axiom of choice for Definition 1 is equivalent for the axiom of choice for Definition 2. The forward direction is trivial; for the reverse direction, assume the axiom of choice for Definition 2 and let $(A_i)_{i\in I}$ be a family of pairwise disjoint sets by Definition 1. Let $J=\{A_i:i\in I\}$ and consider the indexed set $(B_j)_{j\in J}$ defined by $B_j=j$.
I claim $(B_j)_{j\in J}$ is a pairwise disjoint family according to Definition 2. Indeed, suppose $j,j'\in J$ and $j\neq j'$. By definition of $J$, there exist $i,i'\in I$ such that $j=A_i$ and $j'=A_{i'}$. Then $A_i\neq A_{i'}$ and $i\neq i'$ (since $i\mapsto A_i$ is a function), so since $(A_i)_{i\in I}$ satisfies Definition 1, $A_i\cap A_{i'}=\emptyset$. But $A_i=j=B_j$ and $A_{i'}=j'=B_{j'}$, so $B_j\cap B_j=\emptyset$.
Thus by the axiom of choice for Definition 2, there exists a set $C$ such that $|B_j\cap C|=1$ for each $j\in J$. Now for any $i\in I$, let $j=A_i$, so $A_i=B_j$. We thus see that $|A_i\cap C|=|B_j\cap C|=1$. So, this set $C$ also has the required property for the family $(A_i)_{i\in I}$.
Best Answer
With regard to your third question, while I agree with the comments that it's neither necessary nor beneficial to study category theory as a precursor to learning topology, I would make the case for picking up a little category theory along the way, as it becomes relevant (the "just-in-time category theory" approach). For example, when you learn about the product topology and the topological sum, it's appropriate to devote some effort to understanding the universal properties of the product and the coproduct, and when you start learning about the fundamental group or homology, you'll want to learn what a functor is. It may not be strictly necessary in order to understand the topology, but let me give two reasons why picking up this small amount of category theory may be beneficial, in the particular case of products and coproducts.
1. It motivates the construction of the topological product and the topological sum. Sure, you can construct this topology on the disjoint union of some collection of spaces, making it into the topological sum, but why should you do such a thing? Why is the topological sum a thing worth considering? There are many possible answers, but one of them is that the topological sum is the coproduct in the category of topological spaces and continuous functions. Without even worrying about what that means precisely, it says that the topological sum behaves analogously to coproducts in other categories. So, if you believe that the disjoint union of sets (which is the coproduct in the category of sets) or the direct sum of groups or vector spaces (which are the coproduct in the categories of groups or vector spaces, respectively) are interesting or useful, then probably the topological sum is also interesting or useful.
An even better example of this idea is the product of topological spaces. You form the product of spaces by taking the cartesian product of their underlying sets, and then putting a suitable topology on it. Well, there are (at least) two topologies a reasonable person might try, the product topology and the box topology, which are different if you're taking the product of infinitely many spaces. I don't know about you, but when I learned about these for the first time, I was convinced that the box topology was a more sensible choice (why should we restrict all but finitely many of the open sets to be the entire space??). Well, the product topology is the product in the category of topological spaces, and the box topology isn't.
2. More practically, the universal property is useful for producing continuous maps. Have you ever wanted to define a continuous map from the topological sum to some other space? (If you haven't, you will.) More concretely, let's say you want to build a continuous map $$ \coprod_{i \in I} X_i \to Z $$ where the $X_i$'s and $Z$ are some topological spaces. It turns out that the topological sum has exactly the right topology to make this an easy task. All you have to do is produce a continuous map $$ X_i \to Z $$ for each $i$! The universal property of the coproduct says that choosing continuous maps $X_i \to Z$ uniquely determines a continuous map $\coprod X_i \to Z$. I'll grant that you can understand and use this universal property without stating it in the language of category theory, but the same universal property applies to coproducts in any category, so phrasing it in categorical language makes clear similarity in the behavior of objects (such as topological sums and direct sums of vector spaces) that otherwise look completely different.