The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.
That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $\Delta r\to 0,$ the quadratic term in $\Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $\Delta r.$ This is simply because $x^2=o(x),$ in general, as $x\to 0.$
To answer the other questions, which you've numbered and reorganized:
The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.
Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $\underbrace{1×1×1×\cdots×1}_{n} × xy,$ where $n$ is a positive integer.
Thus, you could well think of $2πr\Delta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × \Delta r.$ The annulus with median radius $r$ and width $\Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $\Delta r\to 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.
Consider slicing the cone into many thin coin-like slices. You want to set up the integral that corresponds to adding up the area of the "side wall" of each slice. Which is to say, each slice has a thin circular strip of area that you're after.
You're not taking into account the fact that the sides of the slices are slanted. So for each slice, the width of the strip that you are trying to calculate the area of isn't $\Delta h$, it is $\Delta l$.
Best Answer
You seem to be ignoring the fact that s and r vary as the segment you consider varies. By using the same variable names it appears that you are confusing them to be constants...
Anyway, for a derivation, look at the following figure:
This is a cross-section of the cone.
The area of the strip of width $\displaystyle dh$ that corresponds to $\displaystyle h$ (from the apex) is $\displaystyle 2\pi r \frac{dh}{\cos x}$
Now $\displaystyle r = h \tan x$
Thus $\displaystyle dA = 2 \pi h \frac{\tan x}{\cos x} dh$
Thus the total area
$$= \int_{0}^{H} 2 \pi h \frac{\tan x}{\cos x} dh = \pi H^2 \frac{\tan x}{\cos x} = \pi (H \tan x) \left(\frac{H}{\cos x}\right) = \pi R S$$
Hope that helps.