The principle you have noticed is connected to the system you are using implicitly: spherical, cylindrical and in case of a cube it would be cartesian.
Notice that you need both $r$ and $h$ for the cylindrical system, while only $r$ for the spherical. The remaining coordinates that are gone in the calculation are angle related and since sphere and cylinder are symmetrical this angle components are already counted in. There are no changes over any angle.
Cone and tetrahedron do not fit any of the three systems, as they do not have that form of symmetry, and although your principle applies if you position it all, for example, in the cylindrical system, inevitably you will need somewhere in your calculation a new coordinate component like the lateral height.
In essence, the formula like $\frac{1}{3}\pi r^2 h$ cannot be used since the system you need does not have just $r$ and $h$ coordinates. Contrary to other examples, $r$ and $h$ do not hold a complete information about the geometry of cone or tetrahedron so you could take their derivatives in a simple manner. As we are climbing to a cone, $r$ is changing linearly with $h$ and this is not expressed in the final formula in any particular way.
Again there is nothing wrong with the principle, but deriving it for a cone might not be a direct task. What you can do is expressing the volume in form of an integral (whose result is the standard formula for the volume of cone or tetrahedron) while its form would clearly display the principle you have noticed.
This is not possible directly from the standard volume formula.
To understand the missing link, write the volume of the cone as:
$$V=\frac{1}{3}\pi r^3\tan\alpha$$
where $\alpha$ is the angle of the slant height to the base.
Now write surface area of the cone using the same variables. You have
$$S=\pi r^2(1+\frac{1}{\cos\alpha})$$
and there comes the problem. When you increase $r$ you do not get the same proportional increase in volume and surface area.
However, in which precise way you need to enlarge a cone in order to get its volume and surface area following the same proportions technically having the surface area as a derivative of the volume is not a simple question. But it is definitely possible. The main question is over what we need to express the volume and surface area to have this as you wanted it.
I might suggest expressing the volume and the surface through this two-coordinate system $(u(x),v(x))$ and see if the derivative through $x$ fits your need. (It does not look impossible to have $(r(x),h(x))$ as well, it is just my guessing that $(u,v)$ looks more natural regarding the growth of the cone.)
Let us get to the bottom of this. If you increase all sides by $\Delta x$, as in the picture, it is equivalent to increasing only the height by $\Delta x(1+\frac{1}{\sin{\beta}})$ where $\beta$ is the half angle at the vertex. (Just translate up the internal triangle by $\Delta h$.)
Now instead of derivative we will use equivalent dual number, which is nothing more but introducing a symbol $\epsilon$ so that it is $\epsilon^2=0$. Now finding first derivative of $f(x)$ is the same as calculating $f(x+\epsilon)$ and using cancelation property $\epsilon^2=0$. We replace $\epsilon = \Delta{x}$.
(Uh, now latex is going to kill my fingers but let us do it.)
$$\Delta{h}=\Delta{x}\frac{1}{\sin{\beta}}$$
Now total translation and equivalent height increase is as we mentioned already:
$$\Delta{x'}=\Delta{h'}=\Delta{h}+\Delta{x}=\Delta{x}(1+\frac{1}{\sin{\beta}})$$
Now you can see from the pictures
$$\frac{\Delta{r}}{\Delta{x'}}=\tan{\beta}$$
$$\Delta{r}=\Delta{x}(1+\frac{1}{\sin{\beta}})\tan{\beta}$$
$$V+\Delta{V}=\frac{\pi}{3}(r+\Delta{r})^2(h+\Delta{h'}) = \frac{\pi}{3}(r+\Delta{x}(1+\frac{1}{\sin{\beta}})\tan{\beta})^2(h+\Delta{x}(1+\frac{1}{\sin{\beta}}))$$
From here using cancelation rule and $V=\frac{\pi}{3}r^2h$ we have
$$\Delta{V}=\frac{\pi}{3}(r^2\Delta{x}(1+\frac{1}{\sin{\beta}})+2r\Delta{x}(1+\frac{1}{\sin{\beta}})h\tan{\beta})$$
Since we are interested in partial derivative $\frac{\Delta{V}}{\Delta{x}}$ we have
$$\frac{\Delta{V}}{\Delta{x}} = \frac{\pi}{3}(r^2 + \frac{r^2}{\sin{\beta}}+2rh\tan{\beta} + \frac{2r}{\cos{\beta}})$$
Using obvious properties $\frac{r}{l}=\sin{\beta}$ and $\frac{r}{h}=\tan{\beta}$ we have
$$\frac{\Delta{V}}{\Delta{x}} = \frac{\pi}{3}(r^2 + rl+2rr + 2rl)$$ or as you asked
$$\frac{\Delta{V}}{\Delta{x}} = S = \pi r^2 + \pi rl$$
The above is justifying this parametrization that leads to your answer more or less directly:
$$V=\frac{\pi}{3}r(x)^2h(x)$$ $$r(x)=x(1+\frac{1}{\sin{\beta}})\tan{\beta}$$ $$h(x)=x(1+\frac{1}{\sin{\beta}})$$
$$V=\frac{\pi}{3\tan{\beta}}r(x)^3$$
$$\frac{\partial V}{\partial x}=\frac{\pi}{\tan{\beta}}r(x)^2 r'(x) = \frac{\pi}{\tan{\beta}}r(x)^2 (1+\frac{1}{\sin{\beta}})\tan{\beta}=\pi r^2 + \pi rl=S$$
Adding to the flavor, $x$ is obviously the radius of inscribed circle (of course it is the radius of inscribed sphere when we look at the cube). This is indicating the system we were looking for: the center of the system is the center of inscribed circle and axes are going perpendicularly to the base and side.
P.S. If there is any doubt that this is just a fluke of luck, a cube by the same reasoning has the parametrization $a(x)=2x$ as the radius of inscribed sphere, $x$, is $\frac{a}{2}$.
$$\displaystyle V=a^3$$
$$\displaystyle \frac{\partial V}{\partial x}=3a^2(x)a'(x)=6a^2=S$$
