(This is an expanded version of this).
First of all, the given collection of spaces $\{ Y_j \}_{j \in J}$ can be viewed as a functor $Y: \mathbf{J} \to \mathbf{Top}$, where $\mathbf{J}$ is simply the indexing set $J$ viewed as a discrete category (i.e. one whose objects are $J$ and containing only identity morphisms). In other words, we have $Y(j) = Y_j$ and $Y(\operatorname{id}_j) = \operatorname{id}_{Y(j)}$, which is trivially a functor. The other part of the description of the initial topology is the collection of maps $f_j: X \to Y_j$ where $X$ is a set. More formally, if $U: \mathbf{Top} \to \mathbf{Set}$ is the forgetful functor, then we can also write $f_j: X \to U(Y_j)$ (this takes place in the category $\mathbf{Set}$). As the article mentions, the collection $\{f_j\}$ can be viewed as a cone from $X$ to $UY$. Here comes the definition:
Given categories $\mathbf{C},\mathbf{D}$, and a functor $F: \mathbf{D} \to \mathbf{C}$, a cone from an object $X$ in $\mathbf{C}$ is a natural transformation $\eta: \Delta(X) \Rightarrow F$, where $\Delta: \mathbf{C} \to \mathbf{C^D}$ is the functor mapping each object $Z$ in $\mathbf{C}$ to the constant functor $\Delta(X): \mathbf{D} \to \mathbf{C}$ (which takes each object in $\mathbf{D}$ to $Z$ and each morphism in $\mathbf{D}$ to the identity on $Z$). If $f: V \to W$ in $\mathbf{C}$, then $\Delta(f)$ is the natural transformation $\Delta(f): \Delta(V) \Rightarrow \Delta(W)$ whose component at each object is $f$. For each functor $F$, we define the category of cones $\mathbf{Cone}(F)$ as the comma category $(\Delta \downarrow F)$.
Now, it is easy to see that the collection $f_j: X \to UY(j)$ can be viewed as a cone from $X$ to $UY$, i.e an object in $\mathbf{Cone}(UY)$. In other words, it defines a natural transformation $f: \Delta(X) \Rightarrow UY$ (here $\Delta(X)$ is a functor on the category $\mathbf{J}$, described previously). Similarly, a collection of continuous maps $g_j: Z \to Y_j$ defines an object $\mathbf{Cone}(Y)$. Applying the forgetful functor $U: \mathbf{Top} \to \mathbf{Set}$, we get a forgetful functor $U': \mathbf{Cone}(Y) \to \mathbf{Cone}(UY)$.
Now we are in a position to describe the universal property of the initial topology: as mentioned, we start with an object $f_j: X \to UY$ in $\mathbf{Cone}(UY)$. We will write this as $(X,f)$, where $$f = \{f_j: X \to Y_j\}.$$ Now, a morphism of cones $\tau: (Z,g) \Rightarrow (X,f)$ is simply a morphism $\tau: Z \to X$ such that $f_j \circ \tau = g_j$. Imposing the initial topology (coming from the maps $f_j$) on $X$, we get an object $I(X,f)$ in $\mathbf{Cone}(Y)$. Moreover, the identity map defines a morphism $\varepsilon: U'(I(X,f)) \Rightarrow (X,f)$ (in $\mathbf{Cone}(UY)$), which is universal in the following sense: whenever $\eta: U'(Z,g) \Rightarrow (X,f)$ is another morphism in $\mathbf{Cone}(UY)$, there exists a unique morphism $\xi: (Z,g) \Rightarrow (X,f)$ such that $\eta = \varepsilon \circ \xi$. In other words, if we are given
a topological space $Z$, a collection of continuous maps $g_j: Z \to Y_j$, and a map (a morphism in $\mathbf{Set}$) $\eta: Z \to X$ such that $f_j \circ \eta = g_j$,
then
there exists a (unique) continuous map $\xi: Z \to X$, satisfying $f_j \circ \xi = g_j$ for all $j \in J$, such that $\varepsilon \circ \xi = \eta$.
Of course, since $\varepsilon$ is secretly the identity map on $X$, this implies that $\xi = \eta$ (as set functions, or more accurately, $U(\xi) = \eta$), so all this is saying is that $\xi$ is continuous if $f_j \circ \eta = f_j \circ \xi$ is continuous for all $j \in J$ ($f_j \circ \xi = g_j$, which is the original collection of continuous maps specified above). The converse is of course immediate, since a composite of continuous maps is again continuous. So this highly elaborate categorical language is only saying that
a map $\eta: Z \to X$ is continuous if and only if $f_j \circ \eta$ is continuous for all $j \in J$.
This also implies that imposing the initial topology defines a functor $I: \mathbf{Cone}(UY) \to \mathbf{Cone}(Y)$, and this is right adjoint to the functor $U'$ (and, in fact any right adjoint functor can be described by a similar universal property), but that is another story.
I think the statement as is not quite correct; there is a easy fix however, but first take a look at the following counterexample:
Let $G = \mathbb Z,\ K = \mathbb Z/4$ and $H = 2\mathbb Z$. The projection map $f: \mathbb Z \to \mathbb Z/4$ has the property that for any map $\phi: \mathbb Z \to G'$ such that $2\mathbb Z\subset \ker(\phi)$ (which implies $4\mathbb Z \subset \ker(\phi)$), there is a unique map $\tilde\phi:\mathbb Z/4\to G'$ such that $\phi = \tilde\phi \circ f$. However, clearly the kernel of the map $\mathbb Z\to \mathbb Z/4$ is $4\mathbb Z$, not $H$.
The fix is to require that the map $f:G\to K$ has kernel containing $H$, i.e. $H\subset \ker(f)$. In that case, letting $G' = G/H$ in your statement, we get a unique map $K\to G/H$, and letting $G' = K$ in the universal property of quotient maps, we get a unique map $G/H\to K$. Composition of these two maps $K\to G/H$ and $G/H\to K$ must be identity because of the universal property in your statement. We conclude that $K$ and $G/H$ are isomorphic, and hence $\ker(G\to K) = H$, as desired.
Remark: the proof just given is very typical of the principle "universal objects are unique upto unique isomorphism" in category theory. If you are curious, I recommend skimming the first few pages of Ch.1 of Ravi Vakil's Foundations of Algebraic Geometry (http://math.stanford.edu/~vakil/216blog/FOAGapr2915public.pdf)
Best Answer
1) "Characterized by" means that it is a complete description. A topological space $Y$ is homeomorphic to $X/\sim$ if and only if it satisfies the universal property.
2) A universal property can be used to characterize something if it exists. So you can say that an object $Y$, if it exists, is defined by the following universal property... However, showing existence is a separate issue.
3) Here is a theorem whose proof is quick and clean once you have some categorical properties established. Note that the messy details are hidden in the first line of the proof. If you write out these details, the result is worse than the point-set proof.
Theorem: Let $\sim$ be an equivalence relation on $X$ and let $Y$ be a locally compact Hausdorff space. Then $(X\times Y)/\sim\, \approx \,(X/\sim) \times Y$.
Proof: Since $Y$ is locally compact Hausdorff, the functor $-\times Y$ admits a right adjoint, hence $-\times Y$ commutes with colimits. The universal property of $X/\sim$ is a colimit construction.