Conversion of Additive Noise to Phase Noise
In the following, \$S_n\$ denotes the voltage/current spectrum, as considered in the original question. Considering the spectrum around \$\omega=\omega_0\$,

$$ S_n(\omega) = \frac{1-j}{2} S_{nI}(\omega-\omega_0) + \frac{1+j}{2} S_{nI}(\omega+\omega_0) $$
$$ S_n(\omega_0) = \frac{1-j}{2} S_{nI}(0) + \frac{1+j}{2} S_{nI}(2\omega_0) $$
$$ S_n(\omega_0) = \frac{1-j}{2} S_{nI}(0)$$

since \$S_{nI}(2\omega_0)\approx0\$. The power spectral density is \$\lim_{T\rightarrow 0}\frac{|S_T(\omega)|^2}{T}\$, where \$T\$ is the period and \$S_T\$ is the voltage/current spectrum of a periodic waveform truncated to one period. For energy waveforms (i.e. for the energy spectral density) the limit and division by the period are not required. For ease of notation I use the latter:
$$ |S_n(\omega_0)|^2 = \left|\frac{1-j}{2} S_{nI}(0)\right|^2$$
$$ |S_n(\omega_0)|^2 = \frac{1}2{}\left|S_{nI}(0)\right|^2$$

This derivation assumes that \$S_n\$ is the voltage or current spectrum. I believe that the figure corresponds to having \$S_n\$ denote the power spectrum instead. Thus, it agrees with the figure and the power spectral density of the quadrature components is double that of the RF signal.