Well, I ended up solving this myself after a little more thought. The solution is to convert from the (r, phi, theta) of spherical coordinates to the (x, y, z) of Cartesian coordinates, then swap the z-axis with one of the others before finally converting back to spherical coordinates again.
There is a smooth "spherical coordinates parametrization" $\Phi$ from the closed rectangle $[-\pi, \pi] \times [0, \pi]$ onto the unit sphere $S^{2}$ defined by
$$
\Phi(\theta, \phi) = (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi).
$$
The restriction of $\Phi$ to the open rectangle $(-\pi, \pi) \times (0, \pi)$ is a diffeomorphism to its image. The boundary of the closed rectangle maps to the "International Date Line" $C$, a closed great semi-circle between the north and south poles.
In the complement $S^{2} \setminus C$ of the date line, $\theta$ and $\phi$ are "(smooth local) coordinates". They're not "global", however, because longitude is not continuous and uniquely-defined everywhere on the sphere. (Longitude has a double-valued ambiguity $\pm\pi$ along the date line, and is completely ambiguous at each pole. A bit more correctly, $\theta$ is smooth and single-valued on the complement of $C$, but has no continuous extension to any point of $C$.)
(Incidentally, there's a technical definition of a (local) coordinate system on the sphere, and it turns out there exists no single coordinate system defined on all of $S^{2}$.)
The second assertion is more puzzling (to me). The complement of the equator comprises the northern and southern hemispheres; the complement of the great circle containing the Greenwich meridian comprises the eastern and western hemispheres. Those are presumably the "four domains" the author mentions.
If that's correct, "two different coordinatizations" might mean: A point on the equator is both $0$ degrees north and $0$ degrees south; a point on the Greenwich meridian is both $0$ degrees east and west; a point on the date line is both $180$ degrees east and west.
The main problem is, longitude is in no sense a coordinate at the poles, so I see no geographical sense in which the northern and southern hemispheres are "naturally coordinated". (Certainly there exist perfectly good coordinate systems at the poles, but longitude is not one of the coordinates.)
It's possible the author had in mind that each hemisphere can be expressed mathematically as a graph of a function whose domain is a disk, but even that isn't a trouble-free interpretation:
If the disks are open, the four hemispheres do not cover the entire sphere. (The points on the equator at longitude $0$ and $\pi$ are omitted.)
If the disks are closed, this description fails to satisfy the technical conditions for a "coordinate system" at the boundary of each disk.
Best Answer
To make things simpler I assume $\alpha_0=0$ (the desired value of $\alpha_0$ can always be added/subtracted at the end); furthermore I take the equator at $\delta=0$, so the range of the latitude $\delta$ is the interval $[-\pi/2,\pi/2]$. Your change of coordinates amounts to changing the standard basis $(e_1,e_, e_3)$ of ${\mathbb R}^3$ to the new basis
$$\bar e_1=(\cos\delta_0,0,\sin\delta_0), \quad \bar e_2=(0,1,0),\quad \bar e_3=(-\sin\delta_0,0,\cos\delta_0)\ .$$
It follows that the new coordinates $\bar x_k$ are given in terms of the old ones $x_i$ by the formulas
$$\bar x_1=\cos\delta_0 x_1 +\sin\delta_0 x_3, \quad \bar x_2=x_2,\quad \bar x_3=-\sin\delta_0 x_1+\cos\delta_0 x_3\ .$$
Now we have to express this in terms of the "geographical" quantities $\alpha$, $\delta$, resp. $\bar\alpha$, $\bar\delta$. On the one hand we have
$$x_1=\cos\delta\cos\alpha,\quad x_2=\cos\delta\sin\alpha, \quad x_3=\sin\delta\ ,$$
and on the other hand
$$\bar\alpha=\arg\Bigl({\bar x_1\over\rho},{\bar x_2\over\rho}\Bigr), \quad \bar\delta=\arcsin(\bar x_3)\ ,$$
where $\rho:=\sqrt{\bar x_1^2+\bar x_2^2}$. Putting it all together some simplifications will result.