I'm going to make a not so specific question, since I don't know how to properly state it.
I'm having trouble understanding the final topology on some space $X$ with respect to a family of continuous functions $\{f_i:Y_i\to X\}_{i\in I}$. I believe I have a better understanding of the sum of spaces and of the quotient of spaces (as well, I think I understand better products and subspaces), so my question is: Is the space $X$ endowed with the final topology homeomorphic (or even better, equal) to some construction similar to a quotient of a sum, or a subspace of the quotient of a sum, etc? I think if this were the case, I could better understand how this topology works.
Thanks everyone.
Yes,
Given a family of continuous maps $f_\alpha: Y_\alpha \to X$, if $X = \bigcup_\alpha f_\alpha(Y_\alpha)$, then the topological space defined by the final topology on $X$ is a type of colimit:
$$ X \cong \varinjlim_\alpha f_\alpha(Y_\alpha). $$
It is well known that any finite colimit can be written as a coequalizer of a family of coproducts. That is to say, any such space $X$ is a quotient of sum spaces. The category-theoretic result is the formal dual of Theorem 1 of V.2 of MacLane's Categories for the Working Mathematician. Here's the outline:
Let $Y^\Sigma = \coprod_\alpha Y_\alpha$, and endow $Y^\Sigma$ with the sum topology. There are natural inclusion maps $\iota_\alpha: Y_\alpha \to Y^\Sigma$ coming from the fact that the underlying set of $Y^\Sigma$ is a disjoint union. Put an equivalence relation on $Y^\Sigma$:
$$ \iota_\alpha(x) \sim \iota_\beta(y) \text{ in the case that } f_\alpha(x) = f_\beta(y). $$
Then the quotient space $Y^\Sigma_{/\sim}$ is homeomorphic to $X$. Unfortunately, homeomorphism is the best you can get in the sense that the equivalence relation is defined in terms of the maps $f_\alpha$. This is because we do not generally assume that $X = \bigcup_\alpha f(Y_\alpha)$.
In the case that $X$ is not $\bigcup_\alpha f_\alpha(Y_\alpha)$, then there is a way to recover the "sums and quotients suffice" idea in your question using comma categories and more category theory, but that is considerably more complicated.