Let's first try to develop an idea on how to start the proof. Let's assume for simplicity that $\alpha (x)=x$. We are given that $f:[a,b]\to \mathbb R$ is discontinuous at finitely many points in $[a,b]$.We want to show that $f$ is RS$(\alpha)$ integrable. Now, given any $\epsilon\gt 0$, we want to be able to choose a partition $P_\epsilon$ of $[a,b]$ so that $U(P_\epsilon, f,\alpha)-L(P_\epsilon, f, \alpha)\lt \epsilon$. So let $x_1\lt x_2\lt x_3$ be points in $[a,b]$ where $f$ is discontinuous. Let's "capture" these $x_i$'s in very small intervals. So let's choose $u_i\lt x_i\lt v_i$ such that $v_i-u_i\lt\epsilon/3$ and of course $u_i,v_i$ belong to $[a,b]$. (Note that sum of lengths of these intervals is less than $\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$). Note that the set $K:=[a,b]-\cup_{i=1}^3(u_i,v_i)$ is compact. $P:=\{a\le u_1\lt v_1 \lt u_2\lt v_2\lt u_3\lt v_3\le b\}$ is a partition of $[a,b]$. To take advantage of uniform continuity over $K$, we "refine" $P$ to $P_\epsilon$ by adding points outside $[u_i,v_i]$'s.
Based on the idea developed above, let's proceed with the proof. Let $E=\{x_1,x_2,...,x_m\}\subset [a,b]$ be the points of discontinuity of $f$ in $[a,b]$and assume that $\alpha$ is monotonically increasing. Let $\epsilon\gt 0$ be arbitrary.
Claim: Since $\alpha $ is continuous at $x_i$, we can find $u_i\lt v_i$ such that $\alpha(v_i)-\alpha(u_i)\lt \frac {\epsilon}m. \tag 1$
Proof: Let $\{l_n\}$ (such that $[a,b]\ni l_n\lt x_i$ for all $n\in \mathbb N$) be a sequence converging to $x_i$; and let $\{r_n\}$ be sequence in $[a,b]$ such that $r_n\gt x_i$ for all $n$ and that $r_n$ converges to $x_i$. It follows by continuity of $\alpha$ at $x_i$ that $\alpha (l_n)\to \alpha(x_i)$ and $\alpha (r_n)\to \alpha (x_i)$. It follows that $\alpha (r_n)-\alpha (l_n)\to 0.$ So for some large enough $N$, we must have $\alpha (r_N)-\alpha (l_N)\lt \frac\epsilon m$. Set $v_i=r_N$ and $u_i=l_N$. $∎$
Now, note that $K:=[a,b]-\cup_{i=1}^m(u_i,v_i)$ is compact. $f$ must be uniformly continuous on $K$. There exists a $\delta\gt 0$ such that for $x\in K \land y\in K, |x-y|\lt \delta\implies |f(x)-f(y)|\lt \epsilon$.
So we refine $P$ by adding points $y_j$'s outside $[u_i,v_i]$ such that
None of the $y_j$'s lies in $[u_i, v_i]$ for any $i$.
$y_j-y_{j-1}\lt \delta\iff [y_{j-1},y_j]\cap [u_i,v_i]=\emptyset$
If $j$ is the largest index such that if $y_j\lt u_i$ (where $i$ is the smallest index satisfying this) then $u_i-y_j\lt \delta$; and similarly if $j$ is the smallest index such that $y_{j-1}\gt v_i$ then $y_{j-1}-v_i\lt \delta$.
Let $P':=$ union of these $y_j$'s with $u_i$'s and $v_i$’s. Let $1\le j\le n$ for some appropriate $n$.
Now $U(P',f,\alpha)-L(P',f,\alpha)=\sum_{i=1}^m (M_i-m_i)(\alpha (v_i)-\alpha(u_i))+\sum_{i=1}^n(M_i-m_i)(\alpha(y_i)-\alpha(y_{i-1})$
The first term on RHS is $\le 2M \sum_{i=1}^m (\alpha (v_i)-\alpha (u_i))\lt 2M \epsilon$ (by $(1)$).
The second term on RHS is $\lt \epsilon \sum_{i=1}^n(\alpha(y_i)-\alpha(y_{i-1})\leq \epsilon (\alpha (b)-\alpha (a))$.
So finally, we get: $U(P,f,\alpha)-L(P,f,\alpha)\lt \epsilon ((\alpha (b)-\alpha (a))+2M)$, whence the conclusion follows.
