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.
We have an option to cut the solid of revolution (obtained by revolution of $y = f(x)$ between $x = a$ and $x = b$) into multiple slices in the following manner:
- each slice is a cylinder
- each slice is a section of cone cut by two parallel planes (a frustum of a cone)
Let the desired slicing be done via partition $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ of $[a, b]$. We will apply both the approaches mentioned earlier to calculate the surface area as well as volume of the solid of revolution.
First we deal with volume which has an easier analysis. If we slice the solid as cylinders then the approximation of volume is given by $$V(P) = \pi\sum_{i = 1}^{n}\{f(x_{i})\}^{2}(x_{i} - x_{i - 1})\tag{1}$$ which is a Riemann sum for the integral $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$ and this is the desired volume.
If we slice the solid into frustums of cone we get the approximation of volume as $$V(P) = \frac{\pi}{3}\sum_{i = 1}^{n}\left[\{f(x_{i - 1})\}^{2} + f(x_{i - 1})f(x_{i}) + \{f(x_{i})\}^{2}\right](x_{i} - x_{i -1})\tag{2}$$ which is split into 3 terms and each term is a Riemann sum for $(\pi/3)\int_{a}^{b}\{f(x)\}^{2}\,dx$ so that the desired volume is again $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$
Let's now come to surface area of the solid of revolution. If we slice the solid into cylinders then the surface area is approximated by $$S(P) = 2\pi\sum_{i = 1}^{n}f(x_{i})(x_{i} - x_{i - 1})\tag{3}$$ which tends to $2\pi\int_{a}^{b}f(x)\, dx$.
If we slice the solid into frustums we get the approximation for surface area as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{(x_{i} - x_{i - 1})^{2} + ((f(x_{i}) - f(x_{i - 1}))^{2}}\tag{4}$$ which can be simplified by the use of mean value theorem as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{1 + \{f'(t_{i})\}^{2}}\cdot(x_{i} - x_{i - 1})\tag{5}$$ for some points $t_{i} \in (x_{i - 1}, x_{i})$. This can be split into two sums each of which is a Riemann sum for $\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$ so that the desired surface area is $2\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$.
We see that in case of volume both the approaches give the same answer. But in case of surface area the answers obtained by both the methods are different. Further note that out of the two answers we can easily verify which one is correct by using $y = x, a = 0, b = 1$ so that the solid of revolution is a circular cone. This verification shows that the technique used in equation $(4), (5)$ gives the correct surface area.
The question which OP is asking is this:
Why do both the approaches (using cylinders and frustums) give the same result for volume but different results for surface area?
The reason is simple. Both the sums in $(3)$ and $(4)$ are trying to approximate the surface area of the solid, but there is a huge difference between them namely $$\Delta = 2\pi\sum_{i = 1}^{n}f(x_{i})\left[\sqrt{1 + \{f'(t_{i})\}^{2}} - 1\right](x_{i} - x_{i - 1})$$ and this itself is a non-zero sum unless $f'(x)$ is identically zero. So the approximation $(4)$ is trying to take into account some additional surface area which is left out by sum $(3)$ and this additional part is significant unless $f'(x) = 0$ identically. Hence $(4)$ is a better and correct approximation.
In case of volume both the approximations $(1), (2)$ are Riemann sums for the same integral (but are expressed in slightly different ways).
Best Answer
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.