Here is a proof (without refering to Darboux integrals) of the reverse implication of 6.17 that if $f \alpha'$ is Riemann integrable on $[a,b]$ then $f$ is Riemann-Stieltjes integrable with respect to the integrator $\alpha$ on $[a,b]$ and
$$\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x)\, dx$$
The objective is to show that for any $\epsilon >0$, there exists a partition $P_\epsilon$ such that for any refinement
$$P = \{x_0,x_1,\ldots,x_n:a = x_0 < x_1 < \ldots <x_n = b\}$$
and any choice of intermediate points $\xi_j \in [x_{j-1},x_j]$, we have
$$\left|S(P,f,\alpha,\{\xi_j\})- \int_a^bf(x) \alpha'(x) \, dx\right|= \left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|< \epsilon$$
By the mean value theorem, there exists $\eta_j \in (x_{j-1},x_j)$ for $j=1,2,\ldots,n$ such that
$$\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]= \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j) (x_j - x_{j-1})$$
Thus,
$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|\\= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx + \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j)(x_j-x_{j-1})- \sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})\right|\\ \leqslant \underbrace{\left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|}_{A}+ \underbrace{\sum_{j=1}^n |f(\xi_j)||\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})}_{B}$$
Because $f\alpha'$ is Riemann integrable it follows that there exists a partition $P_A$ such that if $P$ is a refinement, then
$$\tag{1}A= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|< \frac{\epsilon}{2}$$
Since $f$ is bounded, there exists $M>0$ such that $|f(x)| < M$ for all $x \in [a,b]$ and, hence
$$B = \sum_{j=1}^n |f(\xi_j)|[\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < M \sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) $$
It will be shown below that because $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then
$$\tag{2}\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})<\frac{\epsilon}{2M},$$
whence,
$$\tag{3}B < M\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < \frac{\epsilon}{2}$$
Hence, if P is a refinement of the common refinement $P_\epsilon = P_A \cup P_B$, then by (1) and (3) we have
$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|=A+B < \epsilon,$$
and we have proved that $\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x) \, dx$ as was to be shown.
The forward implication can be proved in a similar way.
Proof of (2)
Since $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then for any choice of intermediate points $t_j \in [x_{j-1},x_j]$, we have
$$\tag{4}\left|\sum_{j=1}^n \alpha'(t_j)(x_j - x_{j-1}) - \int_a^b \alpha'(x) \, dx\right|< \frac{\epsilon}{8M}$$
Denoting $M_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$ and $m_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$, by properties of the supremum and infimum there exist points $t_j', t_j''$ such that
$$M_j - \frac{\epsilon}{4M(b-a)} < \alpha'(t_j') \leqslant M_j, \quad m_j \leqslant \alpha'(t_j'') < m_j + \frac{\epsilon}{8(b-a)},$$
and, thus,
$$\sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) < \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1}) -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{2M}\\ = \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx + \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{4M}$$
Hence, since the LHS is nonnegative,
$$\tag{5} \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) \\< \left|\sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx \right|+ \left| \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1})\right|+ \frac{\epsilon}{4M}$$
Using (4), it follows from substitution into (5) that
$$ \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})< \frac{\epsilon}{8M} + \frac{\epsilon}{8M} + \frac{\epsilon}{8M} = \frac{\epsilon}{2M}$$
Noting that $\xi_j, \eta_j \in [x_{j-1},x_j]$ and $|\alpha'(\xi_j)- \alpha'(\eta_j)| \leqslant M_j - m_j$, we obtain the result
$$\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})\leqslant \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})<\frac{\epsilon}{2M}$$