Perhaps this answer will use too much technology. Still, I think it's pretty.
Consider $A_4$ as the group of orientation-preserving symmetries of a tetrahedron $S$. The quotient $X=S/A_4$ is a 2-dimensional orbifold. Let's try to analyse it.
Two-dimensional orbifolds have three different sorts of singularities that set them apart from surfaces: cone points, reflector lines and corners where reflector lines meet. Because $A_4$ acts preserving orientation, all the singularities of $X$ are cone points, and we can write them down: they're precisely the images of the points of $S$ that are fixed by non-trivial elements of $A_4$, and to give $X$ its orbifold structure you just have to label them with their stabilisers.
So what are these points? There are the vertices of $S$, which are fixed by a rotation of order 3; there are the midpoints of the edges of $S$, which are fixed by a rotation of order 2; and finally, the midpoints of faces, which are fixed by a rotation of order 3.
A fundamental domain for the action of $A_4$ is given by a third of one of the faces, and if you're careful about which sides get identified you can check that $X$ is a sphere with three cone points, one labelled with the cyclic group $C_2$ and the other two labelled with the cyclic group $C_3$.
Finally, we can compute a presentation for $A_4$ by thinking of it as the orbifold fundamental group of $X$ and applying van Kampen's Theorem. This works just as well for orbifolds, as long as you remember to consider each cone point as a space with fundamental group equal to its label.
The complement of the cone points is a 3-punctured sphere, whose fundamental group is free on $x,y$. The boundary loops correspond to the elements $x$, $y$ and $xy$. Next, we take account of each cone point labelled $C_n$ by inserting a relation that makes the $n$th power of the appropriate boundary loop equal to $1$. So we get the presentation
$\langle x,y\mid x^2=y^3=(xy)^3=1\rangle$