Loop gain calculation for the buck converter
The transfer function of a CCM-operated buck converter is defined by: 

$$H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}$$

In which the leading term \$H_0=\frac{V_{in}}{V_p}\$. \$V_{in}\$ is the input voltage meaning that the 2nd-order dynamic response of the CCM buck converter is shifting up or down as \$V_{in}\$ changes. \$V_p\$ on the other hand represents the peak voltage of the sawtooth used in the generation of the duty ratio \$D\$ (the leading term has no unit, [V]/[V]). For instance, if a 2-V peak ramp is used, \$V_p=2\$ and corresponds to a 6-dB attenuation. In the example you show, this is a simplified circuit in which the op amp directly drives the duty ratio, assuming \$V_p=1\;V\$ hence the loop gain \$T(s)=H(s)G(s)\$ in dc becomes \$T_0=H_0G_0=\frac{12\;V}{1\;V}1000=12000\$.

One way to overcome the input voltage contribution to the CCM-operated buck converter is to implement feedforward as described in the second edition of my book. Hope this helps clarify things.

Please note that in your expression, this is \$R_{s,CL} = \frac{R_{s,OL}}{1+T}\$ showing how feedback via a high loop-gain reduces the open-loop output impedance (here the dc term only). This open-loop output resistance (dc term) in a voltage-mode buck is \$r_L||R_{load}\$ where \$r_L\$ represents the inductance ohmic loss and is a naturally-low value. In peak current mode control, the open-loop output resistance (the dc term) is mainly dictated by \$R_{load}\$ as the inductor is turned into a voltage-controlled current source.