A function $f$ cannot be both unbounded and Riemann-Stieltjes integrable.
This can be shown by producing an $\epsilon > 0$ such that for any real number $A$ and any $\delta > 0$ there is a tagged partition $P$ with $\|P\| < \delta$ and with a Riemann-Stieltjes sum satisfying
$$|S(P,f,\alpha) - A| > \epsilon$$
Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have
$$|S(P,f,\alpha) - A| = \left|f(t_j)(\alpha(x_j) - \alpha(x_{j-1})) + \sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right| \\ \geqslant |f(t_j)|(\alpha(x_j) - \alpha(x_{j-1})) - \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that
$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|}{\alpha(x_j) - \alpha(x_{j-1})},$$
and it follows that no matter how fine the partition $P$ we have
$$|S(P,f, \alpha) - A| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(P,f,\alpha) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.
By a fixed tag, I'm assuming you mean that for each interval of a partition, you get to pick a specific point of your choice.
If a function satisfies your fixed tag definition then it may still not be Riemann integrable. The function $f : [0, 1]\to [0, 1]$ taking value 1 on rationals and 0 on irrationals is not Riemann integrable. However, it will satisfy your definition and have integral of 1 if your fixed tag is always rational and integral of 0 if your fixed tag is always irrational.
Best Answer
When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.
Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.
Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.
$$f(x) = \begin{cases}0, \quad 0 \leqslant x < 1/2 \\1, \quad1/2 \leqslant x \leqslant 1 \end{cases}\\ G(x) = \begin{cases}0, \quad 0 \leqslant x \leqslant 1/2 \\1, \quad1/2 < x \leqslant 1 \end{cases}$$
With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy
$$1 = L(P,f,G) \leqslant \underline{\int}_0^1 f \, dG \leqslant \overline{\int}_0^1 f \, dG \leqslant U(P,f,G) = 1,$$
the integral exists with $\displaystyle\int_0^1 f \,dG = 1$.
However, for a sequence of partitions $P_n = \left(0,\frac{1}{2n},\ldots, \frac{1}{2} - \frac{1}{2n} , \frac{1}{2} + \frac{1}{2n}, \ldots ,1\right)$, we have $\|P_n\| = 1/n \to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.