I have trouble following the calculation in this answer for Rindler coordinates. In particular, I am not sure how the second term is obtained in this equation:
$$f^x = m\frac{d^2x}{d\tau^2} + ma_0 \frac{1+\frac{a_0 x}{c^2}}{\left(1 + \frac{a_0 x}{c^2}\right)^2 - \frac{v^2}{c^2}}.$$
Ok, let's start with
$$f^x = m\frac{d^2 x}{d\tau^2} + m\Gamma^x_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}\tag{1}\label{eq:1}$$
In Newtonian limit, one can assume that $c\frac{dt}{d\tau} \gg \frac{dx^\mu}{d\tau}$ for $\mu=1,2,3$. So, we have
$$f^x = m\frac{d^2 x}{d\tau^2} + m\Gamma^x_{00} \left(c\frac{dt}{d\tau}\right)^2\tag{2}\label{eq:2}$$
Next using Rindler metric $g$, we compute the Christoffel symbol $\Gamma^x_{00}$
$$\Gamma^x_{00}=\frac12 g^{x\alpha}\left(\frac{\partial g_{\alpha0}}{\partial t}+\frac{\partial g_{\alpha0}}{\partial t}-\frac{\partial g_{00}}{\partial x^\alpha}\right)=-\frac12 g^{xx}\frac{\partial g_{00}}{\partial x}=-\frac12\frac{\partial}{\partial x}\left(-\left(1+ \frac{a_0 x}{c^2}\right)^2\right)=\frac{a_0}{c^2}\left(1+\frac{a_0x}{c^2}\right)\tag{3}\label{eq:3}$$
Inserting into \eqref{eq:2} results in
$$f^x =m\frac{d^2 x}{d\tau^2} + ma_0\left(1+\frac{a_0 x}{c^2}\right)\left(\frac{dt}{d\tau}\right)^2\tag{4}\label{eq:4}$$
From the Rindler line element $ds^2 = -\left(1+ \frac{a_0 x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$, we find that
$$\frac{dt}{d\tau}=\left(\left(1+\frac{a_0x}{c^2}\right)^2-\frac{v^2}{c^2}\right)^{-\frac12}\tag{5}\label{eq:5}$$
Plugging in \eqref{eq:4} gives
$$f^x = m\frac{d^2x}{d\tau^2} + ma_0 \frac{1+\frac{a_0 x}{c^2}}{\left(1 + \frac{a_0 x}{c^2}\right)^2 - \frac{v^2}{c^2}}\tag{6}\label{eq:6}$$
It looks like we are done.
