The fundamental group of a pointed topological space $(X,T,x)$, is the group whose elements are homotopy equivalence classes of paths from $x$ to $x$, and whose composition is the concatenation of such paths.
This induces a function from the class of pointed topological spaces to the class of groups.
Moreover, every continuous function between topological spaces induces a homomorphism between their topological groups.
These two together form a functor $F:\textbf{Top}\to \textbf{Grp}$.
Why is it interesting that this is a functor? What can we do with this fact?
I guess the reason we care about functors in general is that a common way of studying an object $o$ in a category is by studying morphisms $\varphi$ going into/coming out of your object, and so by having a functor we not only get a (possibly simpler) object $F(o)$ but we ALSO get morphisms $F(\varphi)$ going into/coming out of it. Essentially it's preserving how all the objects are related.
Edit: As Joseph Martin points out in the comments to this answer, a functor is also in some sense the "correct" type of mapping between categories. Just like we only care about homomorphisms of groups rather than just plain functions, a functor is something which preserves the essential structure of the object it is being applied to, in this case a category. And just like we have a category $Grp$ where the morphisms are group homomorphisms, we also have $Cat$ where the objects are categories and the morphisms are functors. This is part of the reason why algebraic topologists take great pains in showing that their construction is functorial, because a plain "function" between the object classes forgets too much important structure.
One specific reason toplogists care about functors is that if $F\colon \mathcal{C} \to \mathcal{D}$ is a (covariant in this case) functor and $\varphi \in Mor_{\mathcal{C}}(x, y)$ is an isomophphism, then $F(\varphi) \in Mor_{\mathcal{D}}(F(x), F(y))$ is an isomorphism as well (I leave this as an exercise for you, and to show the analogous statement holds for contravariant functors).
Moreover $\pi_1$ is a homotopy functor, meaning that if $f\sim g$ then $\pi_1(f) = \pi_1(g)$, and hence $\pi_1$ defines a functor on the homotopy category $hTop$ whose objects are the same as $Top$ but whose morphisms are homotopy classes of continuous functions. In particular it follows that a homotopy equivalence of spaces induces an isomorphism of fundamental groups. Therefore if two spaces have non-isomorphic fundamental groups then it's actually functoriality which tells us they are not homotopy equivalent, for example $S^2$ and $T^2$.
Another application of functoriality is to show there is no retraction of the disk to the circle (a corollary of this fact is the Brouwer Fixed Point Theorem). Let $\iota \colon S^1 \to D^2$ denote the inclusion and suppose there is a continuous function $r \colon D^2 \to S^1$ such that $r\circ \iota = id_{S^1}$. Then by functoriality
$$ \pi_1(r) \circ \pi_1(\iota) = id_{\pi_1(S^1)} $$
But since $D^2$ is contractible $\pi_1(D^2) = 0$, so $id_{\pi_1(S^1)}$ must be $0$ since it factors through the $0$ group, which contradicts the fact that $\pi_1(S^1) \cong \mathbb{Z}$. Moreover if you know $\pi_n (S^n) \cong \mathbb{Z}$ then this argument also shows there is no retraction from $D^{n+1}$ to $S^n$.
The non-existence of a retraction could also have been proven using the functoriality of homology, after all $H_n(S^n)$ is much easier to compute. All we need for this argument is some functor $F$ such that $F(S^n)\neq 0$ and $F(\iota) \colon F(S^n) \to F(D^{n+1})$ is the $0$ morphism (so the target category of $F$ should at least have a zero object).
Another Edit: Tobias Kildetoft brought up in the comments a slightly more advanced application of functoriality: using the fact that $\pi_1$ preserves products, functoriality actually implies that $\pi_1(G)$ is abelian for any topological group $G$.
It's more advanced because it requires you know about Group Objects. The idea is that you can define a group as a set $G$ with some functions $m$, $inv$, and $e$ (coming from a one-point set) which satisfy some commutative diagrams involving $G$ and products of $G$. A "group object" in a category $\mathcal{C}$ with finite products is just an object $G$ of $\mathcal{C}$ and some morphisms $(m, inv, e)$ (where now $e$ comes from a terminal object) which make the same diagrams commute. In particular if we have a functor $F\colon \mathcal{C}\to \mathcal{D}$ which preserves products, functoriality implies that $F$ takes group objects in $\mathcal{C}$ to group objects in $\mathcal{D}$.
For example the group objects in $Set$ are groups, and if we add a topology and require the structure functions to be continuous then we see that topological groups are the group objects in $Top$. You might think that every group is a group object in $Grp$ but this is not the case. Remember that the structure on a group object requires morphisms, and for a non-ablian group $G$ the inversion function is not a group homomorphism; therefore if a non-abelian group admits the structure of a group object, it cannot be induced by its internal group operation and must be some external one. But, if you have a group $(G,\cdot)$ that admits a group object structure $(G, m, inv, e)$, then you can use an Eckmann-Hilton argument to show that $m = \cdot$ and in fact the operation is abelian. Indeed the group objects in $Grp$ are exactly the abelian groups, where the group object structure is induced by the internal group structure.
Now the argument that $\pi_1(G)$ is abeliean just goes like this: $G$ is a group object in $Top$, and $\pi_1$ preserves products, so by functoriality $\pi_1(G)$ is a group object in $Grp$, i.e. it's abelian.
A typical way of establishing this result more directly is to use the Eckmann-Hilton argument on the group $\pi_1(G)$, since it has two operations satisfying the interchange law (loop concatenation and point-wise composition). This argument using group objects has the advantage of pushing the formality into a more appropriate setting, so that now we know ANY functor $F\colon Top \to Grp$ that preserves products will take topological groups to abelian groups, and we get an analogous statement for functors between any other two categories with products by computing the group objects.