The laser is providing a periodic electric field and we are anticipating that energy is going to be either lost or gained due to the absorbtion or emission of a phonon. The absorbtion and emission of a phonon is detectable by observing sidebands in the scattered light.
As you've mentioned:
$$\textbf{P} = \alpha \textbf{E}$$
where:
$$\textbf{E} = E_0 cos(\omega_{Laser} t)$$
$$\alpha = \alpha_0 + \frac{d \alpha}{d Q} cos(\omega_{Phonon} t) $$
Therefore, when \frac{d \alpha}{d Q} is non-zero:
$$\textbf{P} = \alpha \textbf{E}$$
$$= [\alpha_0 + \frac{d \alpha}{d Q} cos(\omega_{Phonon} t)] E_0 cos(\omega_{Laser} t)$$
$$= \frac{d \alpha}{d Q} ( cos(\omega_{L} t +\omega_{P} t) + cos(\omega_{L} t -\omega_{P} t)) $$
(ignoring some factors of 2 that will affect the amplitude). Here we see that the scattered light features sidebands that are at frequencies $\omega = \omega_L \pm \omega_P$.
In the case where $\frac{d \alpha}{d Q} = 0$:
$$\textbf{P} = \alpha \textbf{E}$$
$$= \alpha_0 E_0 cos(\omega_{Laser} t)$$
and the scattered light does not feature sidebands.
A polarizability tensor relates the vector components of the applied field to the vector components of the induced polarization by a scaling factor (nine relations in 3-space). For a field $\mathbf F$, polarization $\mathbf p$ and polarizability tensor $\alpha$, one can represent this is through the matrix equation:
$$
\quad
\begin{bmatrix}
p_x\\
p_y\\
p_z
\end{bmatrix}
\
=
\
\begin{bmatrix}
\alpha_{xx} & \alpha_{xy} & \alpha_{xz}\\
\alpha_{yx} & \alpha_{yy} & \alpha_{yz}\\
\alpha_{zx} & \alpha_{zy} & \alpha_{zz}\\
\end{bmatrix}
\
\begin{bmatrix}
F_x\\
F_y\\
F_z
\end{bmatrix}
$$
Tensors have some special (transformation) properties not shared by all matrices, but let's not worry about that now. To see which the tensor elements connect specific field and polarizability elements, just do the matrix multiplication:
$$p_x = \alpha_{xx}F_x + \alpha_{xy}F_y + \alpha_{xz}F_z\\
p_y = \alpha_{yx}F_x + \alpha_{yy}F_y + \alpha_{yz}F_z\\
p_z = \alpha_{zx}F_x + \alpha_{zy}F_y + \alpha_{zz}F_z$$
You may notice that the first index of $\alpha_{jk}$ always corresponds to the polarization direction, and the second index always corresponds to the field direction. For example, the component $\alpha_{xy}$ 'dots' into the $y$ direction of the applied field and results in a polarization in the $x$ direction, while $\alpha_{yx}$ 'dots' into the $x$ field direction and corresponds to a polarization in the $y$ direction. Imagining this 'dotting' with the closest component helps remember which index relates to the field/polarization.
Incidentally, the note above makes a good case for index notation:
$$p_j = \alpha_{jk} F_k.$$
A few more examples may be a helpful reference for relating index notation, matrix notation, and the physical system...
For an isotropic object/medium, the polarization always points in the same direction as the electric field scaled by a constant. This is given by a diagonal matrix:
$$\alpha =
c\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}$$
or in index notation: $p_k = c F_k$.
An object whose polarization is in the same direction as the applied field but is more polarizable in a given direction (say $z$) is given by:
$$p = \begin{bmatrix}c_1 & 0 & 0\\
0 & c_1 & 0\\
0 & 0 & c_2\end{bmatrix}$$
And any polarizability tensor with off diagonal elements result in polarization induced in a different direction than the applied field, as stated above.
Best Answer
The quote you mention is for electric dipole transitions, which are not allowed for homonuclear diatomics. These transitions are for absorption or emission of one photon
You are correct to question absorption and re-emission; this is Raleigh scattering if the photon does not change wavelength (energy). The photon can change energy, though, in Raman scattering. The combined absorption and emssion process is generally very much weaker than allowed emission or absorption of one photon. Modern experiments generally observe Raleigh / Raman scattering with use of lasers, mostly in the visible I think, but no reason that it should not work in the IR as well. One final point is that in Raleight scattering the wavelength of light can change a little bit due to momentum transfer, but the internal energy of the molecule stays the same.
Final point is that some transitions that are not allowed as electric dipole transitions are allowed are electric quadrupole transitions or magnetic dipole transitions, for example. These processes are significantly weaker (less probable) than electric dipole transitions.