Your transformations seem fine. You seem confused about what the transformations are though, since you ask
What does this equation represent?
It represents exactly the same as in Cartesian coordinates. By making the coordinate transformation, you have simply changed the representation of the mathematical structure; not the structure itself.
The equation can be visualized here.
Finding unknown bounds for the cylindrical and spherical cases:
Cylindrical coordinates:
Notice that how far we must go out in the "$r$" direction (in $(r,\theta, z)$ space) is dependent on the value of $z$. If we're at the origin, $r$ has a maximum of $0$ because the vertex is a mere point, but at $z = 1/\sqrt{3}$, $r$ can go all the way out to the edge on the flat top of the cone. Imagine drawing a right triangle with height $z$ and hypotenuse along the outer edge of the cone. Knowing one of the angles of this triangle is $\pi/3$ allows us to "solve the triangle", giving us $\displaystyle r = \frac{\sqrt{3}}{z}$, meaning $r$ is going to run between $0$ and this $z$-dependent value.
Spherical coordinates:
We can take an approach similar to the above to find the bounds for $\rho$. Notice that the upper bound on $\rho$ for given values of $\theta$ and $\phi$ is dependent only on the value of $\phi$. For example, when $\phi = 0$, the maximum value $\rho$ can attain is at its shortest of them all ($0 \leq \rho \leq 1/\sqrt{3}$), but when $\phi = \pi/3$, we have $\rho$ able to attain its longest possible value for the whole cone. Again, draw a triangle as we did above. This time, the triangle will have a height of $1/\sqrt{3}$, and a known angle, allowing you to solve for the hypotenuse, which is $\rho$.
A tip for both of the above:
We are allowed, in this scenario, to rearrange the order of integration. As an example, $\displaystyle \iiint \text{ stuff } \ dr \ dz \ d\theta$ will be the same as $\displaystyle \iiint \text{ stuff } \ dz \ d \theta \ dr$. Let's take advantage of this to the fullest extent possible. In particular, in both cases above, theta will run between $0$ and $2 \pi$, and this is entirely independent of whatever the other variables are doing. So if you think about it, you'll notice that ordering the integrals so that the $d \theta$ is outermost, we'll have:
$$\displaystyle \iiint \text{ stuff }\ dz \ dr \ d\theta = 2 \pi \iint \text{ same stuff } dz \ dr$$
And just like that we have only $2$ integrals to worry about instead of $3$.
Best Answer
There is no general formula. You must have a good understanding of how spherical coordinate works if you are converting to it. You also should have a rough visualization of the surface in your mind. Even a rough $2D$ sketch helps. Then use the limits in cartesian coordinates to find the limits in spherical coordinates.
Now if we work through the example you have written down,
As you mentioned, $x = r \cos \phi \sin \theta, y = r \sin \phi \sin \theta, z = r \cos \theta$.
Now for the cone $z^2 \geq x^2 + y^2$, $r^2 \cos^2 \theta \geq r^2 \sin^2 \theta \implies \theta \leq \frac{\pi}{4}$.
Now bounds of $z$ is $1 \leq z \leq 2$. Given $z = r \cos \theta$, we obtain,
$\sec \theta \leq r \leq 2 \sec \theta$.
Limits of $\phi$ is simply $0$ to $2 \pi$.
So the integral becomes,
$\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_{\sec \theta}^{2 \sec \theta} r^2 \sin \theta \ dr \ d\theta \ d\phi$.