1. $\delta$ and $\Delta$ are just arbitrary constants satisfying $0 < \delta < \Delta$. You may think of them as auxiliary variables that are temporarily frozen during the estimation of the integral. In the end, we will release hold of these variables and take limit as $\delta \to 0^+$ and $\Delta \to \infty$.
2. For $\xi > 0$ and for any $0 < a < b$, we have
\begin{align*}
\left| \int_{a}^{b} f(y)\sin(\xi y) \, \mathrm{d}y \right|
&= \left| \left[ f(y) \frac{1-\cos(\xi y)}{\xi} \right]_{y=a}^{y=b} - \int_{a}^{b} f'(y) \frac{1-\cos(\xi y)}{\xi} \, \mathrm{d}y \right| \\
&\leq \frac{1}{\xi} \left( |f(a)| + |f(b)| + \int_{a}^{b} |f'(y)| \, \mathrm{d}y \right) \\
&\leq \frac{4|f(a)|}{\xi}
= \mathcal{O}\biggl(\frac{1}{a\xi}\biggr).
\end{align*}
So by the Cauchy criterion, the integral $\int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y$ converges uniformly in $\xi \in [\delta, \Delta]$, with the explicit bound
$$ \left| \int_{0}^{b} f(y)\sin(\xi y) \, \mathrm{d}y - \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \right|
\leq \frac{4|f(b)|}{\delta} \tag{1} $$
for any $\xi \in [\delta, \Delta]$ and $b > 0$.
3. Since $(\xi, y) \mapsto \xi^{s-1}f(y)\sin(\xi y)$ is continuous on $[\delta, \Delta] \times [0, b]$, we can interchange the order of integration to get
\begin{align*}
\int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{b} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi
&= \int_{0}^{b} f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y. \tag{2}
\end{align*}
Now,
The left-hand side (LHS) of $\text{(2)}$ converges to $\int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi$ by the uniform convergence established in Item 2. (Even if you are not familiar with the uniform convergence, you can directly employ the bound $\text{(1)}$ to show that the difference tends to $0$ as $b \to \infty$.)
As in Item 2, we can utilize integration by parts to find that
\begin{align*}
\int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi
&= \left[ \xi^{s-1} \frac{1-\cos(\xi y)}{y} \right]_{\xi=\delta}^{\xi=\Delta} - \int_{\delta}^{\Delta} (s-1) \xi^{s-2} \frac{1-\cos(\xi y)}{y} \, \mathrm{d}\xi \\
&= \mathcal{O}(y^{-1}) \quad\text{as $y \to \infty$}.
\end{align*}
Using this, we find that the integrand in the right-hand side (RHS) of $\text{(2)}$ satisfies
$$ f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y = \mathcal{O}(y^{-2}) \quad \text{as $y \to \infty$.} $$
In particular, the integral in RHS of $\text{(2)}$ converges as $b \to \infty$.
Combining these two observations, we get
\begin{align*}
\int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi
&= \int_{0}^{\infty} f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y. \tag{3}
\end{align*}
4. Since the LHS of $\text{(3)}$ converges to the double integral that we started from as $\delta \to 0^+$ and $\Delta \to \infty$, it suffices to prove that the RHS of $\text{(3)}$ also converges to the double integral with the interchanged order of integration. This is explained in detail in the picture, so let me skip this.
