Post History

Let's start with displacement field equation 
$$\vec D = \epsilon_0 \vec E + \vec P$$

We know that $$-\vec \nabla \cdot \vec P = \rho_b$$
Here $\rho_b$ is surface charge density.

$$\implies -\int_V (\vec \nabla \cdot \vec P)\mathrm d\tau=\int_V \rho_b \mathrm d\tau$$
$$\implies \vec P \cdot \hat n = \sigma_b$$
The OP found $\vec D = \sigma_f \cdot \hat n$

$\sigma_b$ is used for surface charge density while $\sigma_f$ is used for free surface charge density if you look at the displacement field equation then you will get

$$\implies\sigma_f = \epsilon_0 \vec E+\sigma_b$$
$$\implies\vec E = \frac{\sigma_f-\sigma_b}{\epsilon_0}$$
$$=\frac{\sigma}{\epsilon_0}$$

>The free charge density serves as a useful simplification in Gauss's law for electricity

> Take a stationary atom and apply an electric field externally to the electron then the free particles align themselves with the field and create an electric dipole $\vec p = q\vec d$ where d is distance between those free particles. 

>If you take a look at electric field generally then you will get the same expression $\vec E = \frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r^2}da\\\\ \implies \vec E = \frac{\sigma}{\epsilon_0}$ so both equation agrees on each other. 

>So polarization is **surface charge density** and electric displacement is **free surface charge density** and direction of polarization and electric displacement depends on electric field. 

Simply, free particles being moved for electric and field and attraction and distraction of other particles