Just as inertial coordinate are radar coordinates for an inertial observer, Rindler coordinates are radar coordinates for an accelerated observer.
Not sure about the readability of what follows.
First, a review of how an inertial observer establishes an inertial coordinate system using her wristwatch and light signals. Suppose ##P## is any event in spacetime. The observer continually sends out light signals, and suppose the light signal that reaches event ##P## left her worldline at time ##t_1## according to her watch. Upon reception at ##P## of this signal, ##P## immediately sends a light signal back to the observer, which she receives at time ##t_2##. If ##x## is the spatial distance of ##P## from the observer's worldline, then, since the light goes out and back, the light travel a distance ##2x## in a time ##\left(t_2 - t_1 \right)##. Thus
$$2x = c \left(t_2 - t_1 )\right.$$
The light spends half the time going out, and half the time coming back. Therefore the time coordinate of event P is the same as the the event on the observer's worldline that is halfway (in time) between the observer's emission and reception events. Consequently, the time coordinate of ##P## is
$$t = \left(t_2 + t_1 \right)/2.$$
It is easy to convince oneself that this operational definition establishes a standard inertial coordinate system.
Note that ##t_1## and ##t_2## are proper times for the observer that sets up the coordinate system.
Assume that an accelerated observer uses the same procedure to establish a non-inertial coordinate system. Consider the case where an observer has constant acceleration. Let ##\left(t , x \right)## be standard coordinates for a global inertial frame. Let the worldline for the accelerated observer be parametrized by her wristwatch (proper) time ##T## so events on her worldline have inertial coordinates
$$\left(t , x \right) = \left(\frac{c}{a} \sinh \left( aT \right) , \frac{c^2}{a} \cosh \left( aT \right) \right).$$
This is one branch of the hyperbola ##c^2 t^2 - x^2 = -\left( c^2 / a \right)^2##.
Now set up a coordinate system for the accelerated observer using the light signal procedure given above for an inertial observer. Let ##P## be an event in spacetime that: receives a light signal that left the accelerated observer at proper time ##T_1##; sends a light signal that the accelerated observer receives at time ##T_2##. The accelerated frame coordinates are
$$\left(t' , x' \right) = \left( \frac{1}{2} \left (T_2 + T_1 \right) , \frac{c}{2} \left(T_2 - T_1 \right) \right).$$
To get a handle on this coordinate system, find what curves of constant ##x'## and curves of constant ##t'## look like in the inertial coordinate system. Let ##Q## and ##R## be the emission and reception events of the accelerated observer, respectively. The inertial coordinates of ##P##, ##Q##, and ##R## are
$$
\left(t_P , x_P \right) = \left(t , x \right)\\
\left(t_Q , x_Q \right) = \left(\frac{c}{a} \sinh \left( a T_1 \right) , \frac{c^2}{a} \cosh \left( a T_1 \right) \right)\\
\left(t_R , x_R \right) = \left(\frac{c}{a} \sinh \left( a T_2 \right) , \frac{c^2}{a} \cosh \left( a T_2 \right) \right)
$$
##QP## lightlike gives
$$
\begin{align}
c \left( t_P - t_Q \right) &= \left( x_P - x_Q \right)\\
ct - \frac{c^2}{a} \sinh \left( a T_1 \right) &= x - \frac{c^2}{a} \cosh \left( a T_1 \right),
\end{align}
$$
which then gives
$$ct - x = \frac{c^2}{a} \left( \sinh \left( aT_1 \right) - \cosh \left( aT_1 \right) \right) = -\frac{c^2}{a} e^{-aT_1} ~~~~~~(1)$$
Similarly, PR lightlike gives
$$ct + x = \frac{c^2}{a} e^{aT_2}~~~~~~(2)$$
Multiplying (1)*(2) gives
$$c^2 t^2 - x^2 = -\left(\frac{c^2}{a} \right)^2 e^{a \left(T_2 - T_1 \right)} = - \left(\frac{c^2}{a} \right)^2 e^{2ax'}.$$
So, curves of constant ##x'## are hyperbolae in the inertial coordinates. The hyperbolae all have asymptotes ##t = x## and ##t = -x##. For ##x>0##, these hyperbolae successively become less sharply curved as x increases.
Dividing (2)/(1) and rearrangement gives
$$t = \frac{e^{2at'} - 1}{e^{2at'} + 1} x$$
So, curves of constant ##t'## are straight lines that pass through the origin of the inertial coordinates. As ##t' \rightarrow \infty##, the lines approach ##t = x##; as ##t' \rightarrow -\infty##, the lines approach ##t = -x##.
The accelerated coordinate system ##\left(t' , x' \right)## only covers the wedge ##x > \left|t \right|## of the global inertial coordinate system, with the halflines ##t = x## and ##t = -x## playing the roles of horizons.
