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).
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 $\mathcal C$.