The Lebesgue-Stieltjes integral is a generalization of the Riemann-Stieltjes integral. Here we illustrate their difference by using a simple example. Let $f = \alpha = 1_{[1/2, 1]}$ on the interval $[0, 1]$.
The Riemann-Stieltjes integral $\int_{[0, 1]} f(t),d \alpha(t)$ do not exist by the following reason. Take the partition $\left\{ 0, \frac{1}{2m}, \frac{2}{2m}, \dots, \frac{m-1}{2m}, \frac{m+1}{2m}, \dots, 1 \right\}$ which skips $1/2$. Let $0 = x_0 < x_1 < \dots < x_n = 1$ be the points in this partition. Note that the Riemann-Stieltjes sum for this partition is $\sum_{i=1}^{n} f(c_i) \left( \alpha(x_i) - \alpha(x_{i-1}) \right)$ where $c_i \in [x_i, x_{i-1}]$ is taken arbitrary. The summand is zero except for the unique interval containing the discontinuity $\frac{1}{2}$, and for that interval we have the term $f(c_i) (1 - 0)$ with the freedom of $c_i \in [(m-1)/2m, (m+1)/2m]$ to choose from. Choosing $c_i$ smaller or larger than $1/2$ wiggles the sum from 0 to 1, no matter how fine the partition is. So the RS integral do not exist. By the very same reason, the RS integral do not exist when $f$ and $g$ share a discontinuous point.
On the other hand, the Lebesgue-Stieltjes integral $\int_{[0, 1]} f(t),d \alpha(t)$ is 1. Simply, the Lebesgue–Stieltjes measure $d \alpha$ is the Dirac (“point”) measure of weight 1 concentrated at the point $1/2$. So $\int_{[0, 1]} f(t),d \alpha(t) = f(1/2)$ for any measurable $f$. Informally, we can say that the LS integral can capture the jump of $\alpha$ at $1/2$ by reading the value of $f$ at exactly $1/2$, no matter whether $f$ and $g$ share the same discontinuity or not. One caveat is that the LS integral can change even if the values of $f$ are fixed almost everywhere. For example, if $f = 1_{(1/2, 1]}$ then $\int_{[0, 1]} f(t),d \alpha(t)$ becomes 0. So we should keep track of all the discontinuities with care.