The author is introducing a definition of a mathematical object, not by constructing it explicitly, but by describing an important property that it satisfies. It is a non-obvious fact that this important property characterizes the object in question "up to unique isomorphism" (which you can read as meaning "uniquely" for the time being).
The universal mapping property is the definition. It is a non-obvious fact that this is a meaningful way to define something.
"The" is shorthand for "unique up to unique isomorphism." I wouldn't worry about this for the time being.
Universal properties can be thought of as a vast generalization of the notion of "largest" or "smallest." In many cases they can be thought of as the "laziest" way to do something. In this case, the free monoid can be thought of as the "laziest" way to turn a set into a monoid. This will be made clearer by a more explicit description of the free monoid (I am assuming that Awodey gives such a description) as the set of words on the elements of $A$.
You might find it helpful to supplement Awodey by reading Lawvere and Schanuel's Conceptual Mathematics.
Let's take the following example:
Fix a field $k$. $\mathcal C :=\ _k\underline{\mathrm{Mod}}$ the category of vector spaces$_{/k}$, $\mathcal D := \underline{Set}$ the category of sets and $F : \mathcal C \to \mathcal D$ the forgetful functor, i.e. $F: V \mapsto V$ as a set and $(f : V \to W) \mapsto (f : V \to W)$ as plain mapping of sets.
Take some set $X$ in $\mathcal D$.
What does it mean for a vector space$_{/k}$ $U(X)$ together with a morphism $L_X : X \to FUX$ to be free over $X$?
Choose a vector space $W$ and a set-theoretic mapping $f : X \to W$.
If $(U(X), L_X)$ is free over $X$ there is a unique morphism $g : U(X) \to W$ such that $F g \circ L_X = f$.
So put more concretely:
There exists a unique $k$-linear mapping $g : U(X) \to W$ such that $Fg \circ L_X = f$. But keep in mind that $Fg$ still is the same mapping - we just "forgot" that is actually is $k$-linear.
$L_X$ is a set theoretic mapping that associates to each $x \in X$ a vector $v_x$ in $U(X)$.
$Fg \circ L_X = f$ just says that $g : v_x \mapsto f(x) \in W$.
The fact that $g$ is unique with this property means: If any $k$-linear map $h$ sending $v_x \mapsto f(x)$, then $g = h$. Hence $g$ is uniquely determined by the images of $v_x$, i.e. $\{v_x | x \in X\}$ is a basis of $U(X)$.
Hence being free over $X$ in this particular case means that $L_X$ identifies $X$ with a $k$-basis of $U(X)$.
This explains the terminology: $U(X)$ is free with basis $L_X(X)$.
As a generaly philosophy behind this you should think of $X$ as some kind of "basis" of $U(X)$ in the sense that any morphism $U(X) \to A$ arises in a unique way by choosing a (potentially simpler morphism) $X \to FA$ and "extending" that to the whole of $U(X)$.
In many cases (like the examples you mentioned), you'll want $\mathcal D = \underline{Set}$ and $F : \mathcal C \to \mathcal D$ some kind of "forgetful" functor, that associates to the objects and morphisms of $\mathcal C$ their "set-theoretic" versions, where one "forgets" the "extra-structure" on $\mathcal C$.
Best Answer
Yes. Notice that this definition works for any functor $U : \mathcal{C} \to \mathcal{D}$; here, it is the forgetful functor $U : \mathbf{Cat} \to \mathbf{Graph}$. The $U$-free object on $X \in \mathcal{D}$ is an object $A \in \mathcal{C}$ with a morphism $i : X \to U(A)$ such that for every morphism $h : X \to U(A')$ there is a unique morphism $f : A \to A'$ with $h = U(f) \circ i$. In other words, it is an initial object in the comma category $X \downarrow U$. This is sometimes also called a universal arrow to $U$. When every object of $\mathcal{D}$ has a universal arrow to $U$, then $U$ is a right adjoint. This is the case here.