The angle of reflection is determined by requiring that the reflected wave matches the phase of the incident wave at all points of the boundary.
When you're working things out in full (usually under the heading of 'Fresnel equations' in EM textbooks), you are basically looking for solutions to Maxwell's equations which consist of
- an incident beam $\mathbf E_\mathrm{in}(\mathbf r,t) = \mathbf E_{0,\mathrm{in}} e^{i(\mathbf k_\mathrm{in}\cdot \mathbf r-\omega t)}$
- a reflected beam $\mathbf E_\mathrm{re}(\mathbf r,t) = \mathbf E_{0,\mathrm{re}} e^{i(\mathbf k_\mathrm{re}\cdot \mathbf r-\omega t)}$
- a transmitted beam $\mathbf E_\mathrm{tr}(\mathbf r,t) = \mathbf E_{0,\mathrm{tr}} e^{i(\mathbf k_\mathrm{tr}\cdot \mathbf r-\omega t)}$
where we implicitly assume that the physical fields are the real parts of the complex fields, and we require that the wavenumbers satisfy the dispersion relation
$$
\omega^2 = c^2 k_\mathrm{in}^2 = c^2 k_\mathrm{re}^2 = \frac{1}{n(\omega)^2} c^2 k_\mathrm{tr}^2
$$
on both sides of the boundary.
Furthermore, if you put the $x$ axis along the boundary and the $z$ axis normal to it, then you can write the incident wavevector as
$$
\mathbf k_\mathrm{in} = k_\mathrm{in} \sin(\theta_\mathrm{in}) \hat{\mathbf{e}}_x - k_\mathrm{in} \cos(\theta_\mathrm{in}) \hat{\mathbf{e}}_z,
$$
so that
$$
\mathbf k_\mathrm{in}\cdot \mathbf r = k_\mathrm{in} \sin(\theta_\mathrm{in}) x - k_\mathrm{in} \cos(\theta_\mathrm{in}) z,
$$
and at the boundary itself
$$
\left.\mathbf k_\mathrm{in}\cdot \mathbf r \right|_\mathrm{boundary}= k_\mathrm{in} \sin(\theta_\mathrm{in}) x
$$
since $z=0$ there.
Similarly, the reflected beam is expected to have a positive $z$ component, so that you can write it as
$$
\mathbf k_\mathrm{re} = k_\mathrm{re} \sin(\theta_\mathrm{re}) \hat{\mathbf{e}}_x + k_\mathrm{re} \cos(\theta_\mathrm{re}) \hat{\mathbf{e}}_z,
$$
and at the boundary you again have
$$
\left.\mathbf k_\mathrm{re}\cdot \mathbf r \right|_\mathrm{boundary}= k_\mathrm{re} \sin(\theta_\mathrm{re}) x.
$$
This is where the phase-matching comes in: the incident beam has a wave dependence as $\mathbf E_\mathrm{in}(\mathbf r,t) \sim e^{i(k_\mathrm{in} \sin(\theta_\mathrm{in}) x - \omega t)}$, while the reflected beam has the functional dependence $\mathbf E_\mathrm{in}(\mathbf r,t) \sim e^{i(k_\mathrm{re} \sin(\theta_\mathrm{re}) x - \omega t)}$. And here is the important thing: the coefficients of those two beams need to be matched using the Maxwell equations at the boundary, but that can only be done at a single point; if we want the two to match everywhere, the functional form needs to match:
\begin{align}
e^{i(k_\mathrm{in} \sin(\theta_\mathrm{in}) x - \omega t)} & \equiv e^{i(k_\mathrm{re} \sin(\theta_\mathrm{re}) x - \omega t)} \quad \forall x,t\\ \ \\
\text{or, in other words,} \quad
k_\mathrm{in} \sin(\theta_\mathrm{in}) & = k_\mathrm{re} \sin(\theta_\mathrm{re}) .
\end{align}
The rest then just falls in by itself: we know from the dispersion relation that $k_\mathrm{in}^2 = \omega^2/c^2 = k_\mathrm{re}^2$, which then requires that $\sin(\theta_\mathrm{in}) = \sin(\theta_\mathrm{re})$ and therefore that $\theta_\mathrm{in} = \theta_\mathrm{re}$.
(Similarly, for the refracted beam, you're also required to have $k_\mathrm{in} \sin(\theta_\mathrm{in}) = k_\mathrm{tr} \sin(\theta_\mathrm{tr})$ through the same phase-matching argument, except that now the dispersion relation requires that $k_\mathrm{tr} = n(\omega)k_\mathrm{in}$, which then implies Snell's law, in the form $\sin(\theta_\mathrm{in}) = n(\omega) \sin(\theta_\mathrm{tr})$.)
Also, one important correction: your source's claim that
From the figure, it is clear that only oscillations normal to the paper can radiate in the direction of reflection, and consequently the reflected beam will be polarized normal to the plane of incidence.
is dead wrong. It is perfectly possible for the charge oscillations in the medium to contain a component in the plane of the page, and it is perfectly possible for the reflected light to contain a polarization component in that plane. There is indeed one and only one angle, known as Brewster's angle, at which the reflected beam is polarized normal to the plane of incidence, but this is a nontrivial consequence of the dynamics encoded in the Fresnel equations, and it is nowhere near obvious.
