If $f \in \mathscr{R}\left(\alpha_1\right)$ and $f \in \mathscr{R}\left(\alpha_2\right)$, then $f \in \mathscr{R}\left(\alpha_1 + \alpha_2 \right)$ and $$ \int_a^b f d\left(\alpha_1 + \alpha_2 \right) = \int_a^b f d\alpha_1 + \int_a^b f d\alpha_2.$$
This is part of Theorem 6.12 (e) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.
Here is my proof:
Let $$\alpha \colon= \alpha_1 + \alpha_2. $$
As $\alpha_1$ and $\alpha_2$ are monotonically increasing functions defined on $[a, b]$, so is $\alpha$.
Moreover, if $P = \left\{ \ x_0, x_1, \ldots, x_n \ \right\}$ is any partition of $[a, b]$, then $$\begin{align} L(P, f, \alpha) &= \sum_{i=1}^n \left( \inf_{x_{i-1}\leq x \leq x_i } f(x) \right) \left[ \alpha \left(x_i \right) - \alpha \left( x_{i-1} \right) \right] \\ &= \sum_{i=1}^n \left( \inf_{x_{i-1}\leq x \leq x_i } f(x) \right) \left[ \alpha_1 \left(x_i \right) - \alpha_1 \left( x_{i-1} \right) \right] \\ & \qquad + \sum_{i=1}^n \left( \inf_{x_{i-1}\leq x \leq x_i } f(x) \right) \left[ \alpha_2 \left(x_i \right) - \alpha_2 \left( x_{i-1} \right) \right] \\ &= L \left( P, f, \alpha_1 \right) + L \left( P, f, \alpha_2 \right), \tag{1} \end{align} $$ and similarly, $$ U (P, f, \alpha) = U \left( P, f, \alpha_1 \right) + U \left( P, f, \alpha_2 \right). \tag{2}$$
Now as $f \in \mathscr{R}\left(\alpha_1\right)$ and $f \in \mathscr{R}\left(\alpha_2\right)$, so for every real number $\varepsilon > 0$ we can find partitions $P_1$ and $P_2$ of $[a, b]$ such that $$ U \left( P_1, f, \alpha_1 \right) - L \left( P_1, f, \alpha_1 \right) < { \varepsilon \over 2 } \ \mbox{ and } \ U \left( P_2, f, \alpha_2 \right) - L \left( P_2, f, \alpha_2 \right) < { \varepsilon \over 2 }. \tag{3} $$
Now if $P$ is any partition of $[a, b]$ such that $P \supset P_1$ and $P \supset P_2$, then we have, for each $j = 1, 2$, $$ L \left( P_j, f, \alpha_j \right) \leq L \left( P, f, \alpha_j \right) \leq U \left( P, f, \alpha_j \right) \leq U \left( P_j, f, \alpha_j \right),$$ and so $$ U \left( P, f, \alpha_j \right) - L \left( P, f, \alpha_j \right) \leq U \left( P_j, f, \alpha_j \right) - L \left( P_j, f, \alpha_j \right). \tag{4} $$ From (3) and (4), we see that, for each $j = 1, 2$, we have $$ U \left( P, f, \alpha_j \right) - L \left( P, f, \alpha_j \right) < { \varepsilon \over 2 }, $$ which together with (1) and (2) yields $$ \begin{align} U (P, f, \alpha) - L (P, f, \alpha) &= \left[ U \left( P, f, \alpha_1 \right) + U \left( P, f, \alpha_2 \right) \right] - \left[ L \left( P, f, \alpha_1 \right) + L \left( P, f, \alpha_2 \right) \right] \\ &= \left[ U \left( P, f, \alpha_1 \right) - L \left( P, f, \alpha_1 \right) \right] + \left[ U \left( P, f, \alpha_2 \right) - L \left( P, f, \alpha_2 \right) \right] \\ &< { \varepsilon \over 2 } + { \varepsilon \over 2 } \\ &= \varepsilon, \end{align} $$ and thus it follows that $f \in \mathscr{R}\left(\alpha_1 + \alpha_2 \right)$.
Finally, $$ \begin{align} \int_a^b f d\alpha &= \int_a^b f d \left( \alpha_1 + \alpha_2 \right) \\ &= \inf \left\{ \ U(P, f, \alpha) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &= \inf \left\{ \ U \left(P, f, \alpha_1 \right) + U \left(P, f, \alpha_2 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &\geq \inf \left\{ \ U \left(P, f, \alpha_1 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} + \inf \left\{ U \left(P, f, \alpha_2 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &= \int_a^b f d \alpha_1 + \int_a^b f d \alpha_2, \tag{5} \end{align} $$ and $$ \begin{align} \int_a^b f d\alpha &= \int_a^b f d \left( \alpha_1 + \alpha_2 \right) \\ &= \sup \left\{ \ L(P, f, \alpha) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &= \sup \left\{ \ L \left(P, f, \alpha_1 \right) + L \left(P, f, \alpha_2 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &\leq \sup \left\{ \ L \left(P, f, \alpha_1 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} + \sup \left\{ L \left(P, f, \alpha_2 \right) \ \colon \ \mbox{ P is a partition of } [a, b] \ \right\} \\ &= \int_a^b f d \alpha_1 + \int_a^b f d \alpha_2. \tag{6} \end{align} $$ From (5) and (6), we can conclude that $$ \int_a^b f d \left( \alpha_1 + \alpha_2 \right) = \int_a^b f d \alpha_1 + \int_a^b f d \alpha_2,$$ as required.
Is this proof sound enough in terms of logic and rigor?
