A universal property of some object $A$ tells you something about the functor $\hom(A,-)$ (or $\hom(-,A)$, but this is just dual). For example, $\hom(R[x],S) \cong |S| \times \hom(R,S)$ is the universal property of the polynomial ring (where $|S|$ denotes the underlying set of $S$). Conversely, we may consider the functor which takes a commutative ring $S$ to $|S| \times \hom(R,S)$ and say that it is a representable functor, represented by $R[x]$. This can be also interpreted as the statement that $R[x]$ is the free commutative $R$-algebra on one generator, see free object for categorical generalizations. Roughly, representing a functor means to give a universal example of, or to classify, the things which the functor describes. This happens all the time in mathematics. Conversely, whenever you have an object $A$, it is interesting to ask what it classifies, i.e. to look at $\hom(A,-)$ and give a more concise description of it. The Yoneda Lemma tells you that all information of $A$ is already encoded in $\hom(A,-)$.
Also, one of the main insights of category theory is that it is very useful to work with morphisms instead of elements. For example, what the quotient ring $R/I$ does for us is not really that we can compute with cosets, but rather that it is the universal solution to the problem to enlarge $R$ somehow to kill (the elements of) $I$. In other words, $\hom(R/I,S) \cong \{f \in \hom(R,S) : f|_I = 0\}$. This makes things like $(R/I)/(J/I) = R/J$ for $I \subseteq J \subseteq R$ really trivial: On the left side, we first kill $I$ and then $J$, which is the same as to kill $J$ directly, which happens on the right hand side. No element calculations are necessary. (On math.stackexchange, I have posted lots of examples for this kind of reasoning.) Quotient rings, quotient vector spaces, quotient spaces etc. are all special cases of colimits.
The universal property of the field of fractions states that $\hom(Q(D),F) \cong \hom(D,F)$, where on the right hand side we mean injective homomorphisms. This says that $Q(-)$ is left adjoint to the forgetful functor from fields to integral domains (in each case with injective homomorphisms as morphisms). This is a special case of localizations. Adjunctions are ubiquitous in modern mathematics. They allow us to "approximate" objects of a category by objects of another category.
So far I have only mentioned some patterns of universal properties, but not answered the actual "philosophical" question "Why are there so many universal properties in math?" in the title. Well first of all, they are useful, as explained above. Also notice that many objects of interest turn out to be quotients of universal objects. For example, every finitely generated $k$-algebra is a quotient of a polynomial algebra $k[x_1,\dotsc,x_n]$. Thus, if we understand this polynomial algebra and its properties, we may gain some information about all finitely generated $k$-algebras. A specific example of this type is Hilbert's Basis Theorem, which implies that finitely generated algebras over fields are noetherian. Perhaps one can say: Universal objects are there because we have invented them in order to study all objects.
As hinted in the comments by Menezio the kernel a map $f\colon M\to N$ is the equalizer of $f$ and the trivial map $0$. This is a natural requirement as we expect from the kernel that every element $x\in\ker f$ is taken to the $0$-element in $N$ after pre-composing with the inclusion $\iota\colon\ker f\to M$. Equivalently arrow-wise: we expect $f\circ\iota=0$, i.e. that pre-composing with $\iota$ equals the trivial map.
Now, the equaliser ${\rm eq}(f,0)$ consist of an object $\ker f$ and a map $\iota\colon\ker f\to M$ such that $f\circ\iota=0\circ\iota=0$ (as composing with the trivial map is always trivial) and is universal as such. That is, for every map $k\colon K\to M$ with $f\circ k=0~\circ k=0$ there exists an unqiue map $\tilde k\colon K\to\ker f$ such that $\iota\circ\tilde k=k$.
To put it otherwise: the kernel of a map $f\colon M\to N$ is universal among maps $g\colon K\to M$ such that $f\circ g=0$, which might sound like the more usual definition. Explicitly
The kernel of $f\colon M\to N$ is the unique par $(\ker f,\iota)$, where $\iota:\ker f\to M$ and $f\circ\iota=0$. This pair is universal in so far that for every $f\circ g=0$, with $g\colon K\to M$, there exists a unique $\tilde g\colon K\to\ker f$ such that $\iota\circ\tilde g=g$.
Best Answer
Products in many categories are said to possess a universal mapping property- formally, a product of two objects (say $A$ and $B$), is another object (say $P$) along with two arrows: $p_1:P \to A$ and $p_2:P \to B$ such that if $C$ is any other object (in our category), along with any pair of arrows, $f_1:C \to A, f_2: C \to B$, then there exists a UNIQUE arrow $\phi:C \to P$ such that:
$p_1\circ\phi = f_1\\p_2\circ\phi = f_2.$
The "baby example" of this construction is the cartesian product in the category $\mathbf{Set}$. Indeed, if we set, for two sets $A,B$:
$P = \{(a,b): a \in A, b \in B\}$, and take:
$p_1((a,b)) = a\\p_2((a,b)) = b,$
Then given any pair of maps $f_1:C \to A, f_2:C \to B$ for an abitrary set $C$, we can let:
$\phi(c) = (f_1(c),f_2(c))$, for any $c \in C$, and it's clear we have:
$(p_1\circ\phi)(c) = p_1(\phi(c)) = p_1(f_1(c),f_2(c)) = f_1(c)$ for ALL $c \in C$, and similarly:
$(p_2\circ \phi)(c) = f_2(c)$.
So certainly the map $\phi$ exists (for "this product"). On the other hand, if we have a map $\psi: C \to A \times B$ such that:
$p_1 \circ \psi = f_1$ and $p_2\circ \psi = f_2$, then from $p_1\circ \psi = f_1$ we know that:
$\psi(c) = (f_1(c),-)$, and similarly we know that $\psi(c) = (-,f_2(c))$, that is: $\psi$ has to be $\phi$, so $\phi$ is unique.
Now it should be pretty clear that this isn't "the only product" we can make: for example, we could take:
$P' = B \times A$ with $p_1:P' \to A$ given by $p_1'((b,a)) = a$, and similarly for $p_2'$.
In this case, we have two unique maps $\phi: P \to P'$, and $\phi': P' \to P$, and since $\phi((a,b)) = (b,a)$ and $\phi'((b,a)) = (a,b)$ fulfil the requirements, they must be the maps the universal property describes. Here, it's clear these are isomorphisms (in $\mathbf{Set}$, isomorphisms are just bijections), as these maps are inverses of each other.
The universal property invoked here is often paraphrased as: "the product map (what we are calling $\phi$, and is often written $f_1 \times f_2$) factors through the projections (what we are calling $p_1,p_2$)".
Now, it's "intuitively clearer" to regard the cartesian product of sets to be "the product" of $A$ and $B$. Note how the emphasis is on the elements in each ordered pair. In the construction above, the emphasis is on the mapping $\phi$, and the projection maps $p_1,p_2$, we don't even "look inside" the sets to see who lives there.
There is a similar factorization happening in your field of fractions example: any embedding $f$ of $D$ in any field $F$ factors through the embedding $i: D \to Q$. This captures (in this case) the sense of $Q$ being the "smallest" field we can insert $D$ into.