Just to clarify that if $\displaystyle\int|g|=0$, then $g=0$ a.e. and $\gamma\in[\alpha,\beta]$ can be taken as arbitrary.
If it were not, then $\gamma=\dfrac{\displaystyle\int f|g|}{\displaystyle\int|g|}$ would be a candidate. And we know that $(\cdot)\geq(\cdot\cdot)$ implies that $\displaystyle\int(\cdot)\geq\int(\cdot\cdot)$ no matter $(\cdot),(\cdot\cdot)$ are positive or negative or not.
If you can establish that the Riemann-Stieltjes integral $\int_a^b\alpha \, d\alpha$ exists by hypothesis or otherwise, then it is straightforward to show that
$$I= \int_a^b \alpha \, d\alpha = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$
Note that for any partition $a = x_0 < x_1 < \ldots < x_n = b$ we have
$$\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]=\frac{1}{2}\sum_{j=1}^n[\alpha^2(x_j) - \alpha^2(x_{j-1})] =\frac{1}{2}\sum_{j=1}^n[\alpha(x_j) + \alpha(x_{j-1})][\alpha(x_j) - \alpha(x_{j-1})] \\ = \frac{1}{2}\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})] +\frac{1}{2}\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})] $$
Thus,
$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|\leqslant \\ \frac{1}{2}\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right| +\frac{1}{2}\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right|$$
Note that the sums on the RHS are both Riemann-Stieltjes sums. Since the Riemann-Stieltjes integral, $I$, exists, for any $\epsilon > 0$ there exists a partition such that
$$\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right|< \epsilon, \\\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right| < \epsilon$$
Hence, for any $\epsilon > 0$ we have
$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|< \epsilon,$$
and it follows that
$$I = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$
If $\alpha$ is increasing then it might have jump discontinuities in which case the RS integral of $\alpha $ with respect to $\alpha$ cannot exist. To carry on we must assume that $\alpha$ is continuous. Then it is not difficult to show that $\int_a^b\alpha \, d\alpha$ exists using both continuity and the fact that $\alpha$ is increasing (or more generally because it is of bounded variation).
Best Answer
Since $f$ is RS-integrable with respect to $\alpha$, for any $\epsilon > 0$ there exists a partition $P_\epsilon$ of $[a,b]$ with points $ a = y_0 < y_1 < \ldots < y_m = b$ such that if $P$ is a refining partition with points $ a = x_0 < x_1 < \ldots < x_n = b$ , then for any Riemann-Stieltjes sum
$$S(P,f,\alpha) = \sum_{j=1}^n f(\xi_j)( \, \alpha(x_j) - \alpha(x_{j-1})\, )$$
where $\xi_j \in [x_{j-1}, x_j]$ are any intermediate points, we have
$$\left|S(P,f,\alpha) - \int_a^b f \, d\alpha \right| < \epsilon$$
Since $g$ is strictly increasing and continuous there exists a continuous inverse $g^{-1}:[a,b] \to [c,d]$ such that $g^{-1}(a) = c$ and $g^{-1}(b) = d$
This gives a partition $P'_\epsilon = g^{-1}(P_\epsilon) = \{g^{-1}(y_0), g^{-1}(y_1), \ldots , g^{-1}(y_m)\}$ such that, for any $P$ that refines $P_\epsilon$, the partition $P' =g^{-1}(P)= \{g^{-1}(x_0), g^{-1}(x_1), \ldots , g^{-1}(x_n)\} $ is a refinement of $g^{-1}(P_\epsilon)$. There is a one-to-one correspondence between all partitions $P$ of $[a,b]$ and all partitions $P'$ through the mapping $g^{-1}$.
Also by monotonicity, for any intermediate point $\xi_j \in [x_{j-1},x_j]$ the corresponding point is $g^{-1}(\xi_j) \in [g^{-1}(x_{j-1}), g^{-1}(x_j)]$.
Note that $f(\xi_j) = f(g(g^{-1}(\xi_j)))$ and $\alpha(x_j) = \alpha(g(g^{-1}(x_j)))$,
Thus,
$$\left|S(P',f \circ g,\alpha \circ g) - \int_a^b f \, d\alpha \right|=\left|S(P,f,\alpha) - \int_a^b f \, d\alpha \right| < \epsilon$$
This proves that the composition $f \circ g$ is Riemann-Stieltjes integrable with respect to the composition $\alpha \circ g$