Universal property of a free object I am given the following Universal Property of a Free Object:

Given categories $\mathcal C, \mathcal D$ and a functor $F: \mathcal C \rightarrow \mathcal D$, then for $X \in Ob(\mathcal D)$:
We say an object $U(X) \in Ob(\mathcal C)$ together with a morphism $L_X \in \hom_{\mathcal D}(X, FU(X))$ is Free on X if:
$\forall A \in Ob(\mathcal C)$ and $f \in \hom_{\mathcal D}(X, FA)$, there is a unique morphism $g \in \hom_{\mathcal C}(U(X), A)$ such that $F(g)L_x = f$

This result was given in a course that did not require Category Theory, and I have not done any Category Theory so I am struggling to understand the meaning behind this.
Intuitively, what does it mean for $(U(X), L_X)$ to be free on $X$? If this is too vague and abstract a question without context, what does it mean if the categories here are modules, algebras, rings, or groups? (These tend to be the main focus of the course).
 A: 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 $\mathcal C$.
