I like this question; what you're trying to understand is important to understand.
In this answer I'll be talking loosely about infinitesimal quantities linear or quadratic in $\mathrm dx$; I think this is the best way to get a feel for this sort of thing, but similar arguments could also be presented more rigorously.
Basically, the reason is that in the case of the surface area, the effect from the slope is linear in $\mathrm dx$, whereas in the case of the volume, it's quadratic in $\mathrm dx$. Thus we can neglect it in the limit $\mathrm dx\to0$ in the latter case but not in the former.
Let's see what happens if we take the slope into account in adding up the volume of slices of the solid of revolution generated by a function $f(x)$ rotated around the $x$ axis. As you say, after the cylindrical volume $\pi f^2\mathrm dx$ the next order of approximation would be a cone, or more precisely a conical frustum, corresponding to a linear approximation to the function. The volume of such a frustum between $x$ and $x+\mathrm dx$ would be
$$\begin{eqnarray}
\frac13\pi\mathrm dx\left(f(x)^2+f(x)f(x+\mathrm dx)+f(x+\mathrm dx)^2\right)
&\approx&
\frac13\pi\mathrm dx\left(3f(x)^2+3f(x)\mathrm dx\right)
\\
&=&
\pi\mathrm dx\left(f(x)^2+f(x)\mathrm dx\right)\;,
\end{eqnarray}
$$
which differs from the cylindrical volume by the second term, which contains one more factor of $\mathrm dx$ than the first one and therefore vanishes in the limit.
By contrast, for the surface area, taking into account the slope leads to a surface element $2\pi f(x)\sqrt{1+f'(x)^2}$, whereas not taking it into account would lead to just $2\pi f(x)$, the surface area of a cylindrical slice. Here we don't have two terms with one negligible and dominated by the other, but an additional factor that survives the limit.
You can also try to picture this geometrically. Think of a conical slice and the corresponding cylindrical slice, and imagine shrinking their width. As you shrink, the portion of volume in that little extra bit on the boundary becomes negligible compared to the bulk of the slice -- whereas the bulk only shrinks with the width, the extra bit shrinks both with the width and with the vertical deviation, which is the slope times the width, so it shrinks quadratically while the bulk shrinks linearly. For the surface, there's no such effect, since there's no "bulk" of the surface; all of the surface is at the boundary, and tilting it by the slope makes all of it larger, not just a small portion that becomes negligible in the limit.
One way to gain the intuition behind this is to look at what happens in 2 dimensions. Here, rather than surface area and volume, we look at arc length and area under the curve. When we want to find the area under the curve, we estimate using rectangles. This is sufficient to get the area in a limit; one way to see why this is so is that both the error and the estimate are 2-dimensional, and so we aren't missing any extra information.
However, the analogous approach to approximating arc length is obviously bad: this would amount to approximating the curve by a sequence of constant steps (i.e. the top of the rectangles in a Riemann sum) and the length of this approximation is always just the length of the domain. Essentially, we are using a 1-dimensional approximation (i.e. only depending on $x$) for a 2-dimensional object (the curve), and so our approximation isn't taking into account the extra length coming from the other dimension. This is why the arc length is computed using a polygonal approximation by secants to the curve; this approximation incorporates both change in $x$ and change in $y$.
Why is this relevant to solids of revolution? Well, in essence, the volume and surface area formulae are obtained by simply rotating the corresponding 2-dimensional approximation rotated around an axis, and taking a limit. If it wouldn't work in 2 dimensions, it certainly won't work in 3 dimensions.
Best Answer
I'm not sure about your setup, but in any case you should be integrating in a direction perpendicular to the $r$ in your cross section.
I'll use a setup that's hopefully what you had in mind: the tip of the cone is at the origin, and the cone is lying on its side. Suppose the radius of the base is $r$ (this is the $y$ value at the end of the triangle you're rotating) and the height is $h$ (this is the $x$ value at the end of the triangle).
Then, consider a circular cross section at some point $x$. The radius is the corresponding $y$ value, which by similar triangles is $\frac{rx}{h}$. So, the area is $$ A(x) = \frac{\pi r^2}{h^2} x^2, $$ and the volume is $$ V = \int_0^h \frac{\pi r^2}{h^2} x^2\ dx = \frac{\pi r^2 h^3}{3h^2} = \frac{1}{3} \pi r^2h. $$ Note the cone lies on its side, so the $x$ values we integrate over range from $0$ to the "height" of the cone, $h$.