I looked up the definition of an infinite product of (finite) groups on Wikipedia. It says that the product of the groups $G_i, i\in I$, as a set, is the set of functions $$ f: I \to \bigcup_{i\in I}G_i $$ satisfying some things.
My question is, how is the union defined? The union makes sense if the groups are subgroups of some large group, but otherwise I don't understand how this is defined.
On
This is the ordinary union of sets.
One thing of the "some things" your functions should satisfy is that $f(i) \in G_i$. A very clean and useful way to think about this is the following. First, think about a finite set $I = \{1, \dots, n\}$. Then such an $f$ can be described by a tuple $(g_1, \dots, g_n)$, where $f(i) = g_i$ for $1 \leq i \leq n$. This agrees with the standard interpretation of what a product is, doesn't it?
Next, think about $I = \mathbb N$. Then we have practically the same thing, but now our tuple describing our function is an infinite tuple $(g_1, g_2, \dots)$.
Finally, and this is where it kind of gets important that we have a formal notion, if $I$ is uncountable the whole tuple visualization thing breaks down. How does a tuple look like whose indices can be, say, any real number in $\mathbb R$? I don't know. So instead of trying to define what uncountably big tuples are, the disjoint union is used as a hack to get around this.
But honestly, thinking about "uncountably long tuples" probably works in most situations.
By definition, $\bigcup_{i\in I}G_i$ is just the set of all $x$ such that $x\in G_i$ for some $i\in I$. You just literally take all the elements of all the $G_i$ and put them together in one big set. There is no reason to think that this union has a natural group structure or anything like that, but that's fine; it's just a bare set of elements, and that's all you need in order to be able to talk about functions $I\to \bigcup_{i\in I}G_i$.