Imo, the point of the Riemann-Stieltjes integral is not that "We can now integrate more functions, which makes us happy." (btw. I'm doubtful this is actually true). The point is, that we now have a more general way of expressing certain things as an integrals, or a clearer view of what an integral naturally should be.
For example: We may think of the expression $\int_a^b \rho \, dx$ as the total mass of some segment $[a,b]$, where $\rho: [a,b] \to \mathbb R$ is the mass density on $[a,b]$. In this situation, the Riemann integral is sufficient.
But what the Riemann integral allows us to do may not be sufficient in some cases. Sometimes, we really would like to talk about point-like objects, such as point-masses. Intuitively, the density $\rho$ of a point-mass situated at $x=0$, say, should be zero everywhere, except at $0$. At the same time, we would like to have
$$\int_{\mathbb R} \rho \, dx = \mathrm{mass}.$$
Now, with a Riemann integral, we can never achieve this. There simply is no function $\rho$, such that $\rho(x)=0$ for $x\ne 0$, and yet $\int_{\mathbb R}\rho \, dx \ne 0$.
Bring in the Riemann-Stieltjes integral: Let $F$ be given by $F(x) = 0$, if $x<0$, and $F(x) = \mathrm{mass}$, if $x\ge 0$. Then $F(x)$ is the total mass on $(-\infty, x]$. Now, $F$ does not have a density of the form $\rho\, dx$, but it does have a density $dF$ in the R-S sense. And this does what we want: We have
$$\int_{-\infty}^x \, dF = \begin{cases} 0 & x< 0\\ \mathrm{mass} & x\ge 0.\end{cases}$$
So, using R-S integrals, we have gained a generalized notion of "density". In fact, we learn from this that when speaking about the density of something, we should not think of it in terms of a function $\rho$, but rather in terms of an expression like $\rho dx$, or $dF$. Said differently, we see that the Riemann integral, or the $dx$ density, is just a very special way of measuring something, which pertains to certian situations but not to others. There are situations, where measurement should be done in a way which is not captured by the Riemann integral.
This realization may have been the first step in the direction of the notion of a measure. A concept which underlies much of modern analysis.
It is in this way, that the Riemann-Stieltjes integral generalizes the Riemann integral.
Best Answer
The easiest example to use is the indicator function of the rationals. It takes the value 1 for rational numbers and the value 0 for irrational numbers. Since both the rationals and the irrationals are dense in $\mathbb{R}$, the highest value in every interval of the partition is 1 and the lowest is 0.Take this function on the interval $[0, 1]$.