When we get Riemann-Stieltjes integral becomes standard Riemann integral which calculates area under the curve.
We have that
$$
s(f,\alpha,P)=\sum_{k=1}^nm_k\Delta\alpha_k
\ \text{ and }\ S(f,\alpha,P)=\sum_{k=1}^nM_k\Delta\alpha_k
$$
and that lower $\int_{a}^{b}f(x)d\alpha = \sup_P s(f,\alpha,P)$ and upper $\int_{a}^{b}f(x)d\alpha = \inf_PS(f,\alpha,P)$. We also have that $\Delta\alpha_k=\alpha(x_k)-\alpha(x_{k-1})$.
So when $\alpha(x) = x$ we get $\Delta\alpha_k=x_k-x_{k-1}$ in other words we get Riemann integral.
But if we have for example $\alpha(x) = e^x$ then it is clear that $\Delta\alpha_k=e^{x_k}-e^{x_k-1}$. It is clear that for any function $f$ we have that
$$s(f,x,P) < s (f,e^x,P)
\ \text{ and }\ S(f,x,P) < S(f,e^x,P).
$$
This implies that lower $\int_{a}^{b}f(x)d\alpha = \sup_P s(f,x,P) < \int_{a}^{b}f(x)d\alpha = \sup_P s(f,e^x,P)$ and same thing is for upper integral.
So if we have $f=(x)$ and $\alpha(x) = x$ we get Riemann integral which calculates area under the curve and when $\alpha(x) = e^x$ we get Riemann-Stieltjes which is greater then Riemann integral for the same function $f=(x)$. This begs the question what does Riemann-Stieltjes integral calculate when $\alpha(x) \neq x$?
Best Answer
One uses Riemann-Stieltjes when integrating some function along a curve with respect to the curve length.
When looking for bounded functionals on the space of continuous functions (in sup norm), one again encounters Riemann-Stieltjes integrals, here one can visualize them as encoding a density function that is not expressible as derivative. For instance, one can construct the Dirac delta as Riemann-Stieltjes with the unit jump (Heaviside function) as integrator.
Your example comparing $x$ and $e^x$ requires a non-negative $f$, so "any" does not work for $f$. Also, $\int_a^b f(x)\,d(e^x)=\int_a^b f(x)\,e^x\,dx$, so the monotonicity of the Riemann integral already fully justifies the inequality of the integrals.