TL;DR: ultimately you do study cross-sections of other cone-like shapes, you just don't know about it unless you've studied some algebraic geometry.
The cross-sections of a cone lie in perspective. Imagine you are sat at the origin in $\mathbb{R}^{3}$ and you look out along a ray (=half-line) which lies inside the cone. Fix a plane $\Pi$ which does not go through the origin in $\mathbb{R}^{3}$ and look at the conic section $C$ given by $\Pi.$ From your point of view, the conic section $C$ looks like some vaguely oval shape. Now rotate the plane $\Pi$ about a fixed point on $C$; amazingly enough, your point of view doesn't change! Your vaguely oval shape stays completely still throughout the entire process. Even though, from an extrinsic perspective, the conic section $C$ is changing dramatically during the process - maybe at first it was a circle, and then as $\Pi$ rotates, it becomes an ellipse, then very briefly a parabola, and then a hyperbola - the whole time, what you see from the origin looks the same.
Roughly speaking, the study of this phenomenon used to be known as projective geometry. Modern projective geometry has its roots in this, but has grown far outside these bounds.
Suppose you give me a polynomial $f(x,y)$ in two variables; then the equation $f(x,y)=0$ defines a set of points in $\mathbb{R}^{2},$ and this set of points forms a curve. The curve might be a "conic section", like in the case $f(x,y)=x^{2}+y^{2}-1,$ or it might not be, like in the case $f(x,y)=y^{2}-x^{3}-x^{2}+1.$
Here's a neat trick: make a new polynomial $F(x,y,z),$ this time in three variables, as follows: $F(x,y,z)=f(x/z,y/z)\cdot z^{\deg{f}}.$ Let's look at the two examples I mentioned above, to get a feel for this. If $f(x,y)=x^{2}+y^{2}-1,$ then
$$F(x,y,z)=f(x/z,y/z)\cdot z^{2}=z^{2}\left(\frac{x^{2}}{z^{2}}+\frac{y^{2}}{z^{2}}-1\right)=x^{2}+y^{2}-z^{2}.$$ In the same way, if $f(x,y)=y^{2}-x^{3}-x^{2}+1,$ then $$F(x,y,z)=z^{3}\left(\frac{y^{2}}{z^{2}}-\frac{x^{3}}{z^{3}}-\frac{x^{2}}{z^{2}}+1\right)=y^{2}z-x^{3}-x^{2}z+z^{3}.$$ By construction, what we get at the end is always a homogeneous polynomial, that is, every monomial term in the polynomial is of the same fixed degree (in the above examples, $2$ and $3$, respectively).
Why is this cool? Consider the set of points in $\mathbb{R}^{3}$ satisfying $F(x,y,z)=0.$ What does this look like? In the case of the circle $f(x,y)=x^{2}+y^{2}-1,$ we got $F(x,y,z)=x^{2}+y^{2}-z^{2},$ and $\{(x,y,z):x^{2}+y^{2}=z^{2}\}$ is exactly the standard cone in $\mathbb{R}^{3}$! The "conic section" $x^{2}+y^{2}=1$ is just what you get when you intersect this surface with the plane $\Pi=\{z=1\}.$ And in fact, more generally, if you give me any homogeneous polynomial $F(x,y,z),$ then the set $\{(x,y,z):F(x,y,z)=0\}$ will be a cone over the curve $f(x,y)=F(x,y,1)=0.$ Make no mistake: this "cone" doesn't look like the kind of cone you may be used to; it will be weirdly shaped, so that the intersection with the plane $\Pi=\{z=1\}$ will give back the curve $F(x,y,1)=0.$
What does this mean? It means that whenever you consider an algebraic curve $f(x,y)=0$ in in the plane (and I wager that you do this a lot, whether or not you realize it), you are in fact already considering a "conic" section; it's just the cone is not quite what you are expecting! But it does have the same crucial "perspective" property that I point out before.
In algebraic geometry, you would study the homogenization $F(x,y,z)=0,$ but the modern viewpoint is not to think of this as a surface in space, but instead as a curve in the projective plane $\mathbb{P}^{2}$: each line through the origin in $\mathbb{R}^{3}$ is represented by a point in $\mathbb{P}^{2}.$ I'll leave it to you to imagine a line through the origin rotating around in space, and sometimes it is contained, as a subset, in the surface $F(x,y,z)=0$ (think of the special case of the ordinary cone $x^{2}+y^{2}=z^{2}$ with which you're already familiar), but usually it only intersects this surface at the origin; those times when it is contained in the surface are the "points" on the "curve" $F(x,y,z)=0$ in $\mathbb{P}^{2}.$
So why the focus on classical conic sections?
This has a simple answer: because they are the easiest case! They are one of the few truly accessible cases in algebraic geometry, in the sense that it is very easy to classify them: you have circles, hyperbolas, and ellipses (plus some degenerate cases). With more general "cones", this classification is not always so easy! And moreover, there is a fair amount to be said, many ways to say those things, and the maths relates to lots of other things (besides geometry, the study of conic sections is by definition also the study of quadratic forms).