It's straightforward to show that
@vanhees71's definition #105 follows from the basic
convention of
Einstein-Poincaré simultaneity [a special case of Reichenbach simultaneity for ##\epsilon = \dfrac{1}{2}##]. If light is emitted from a point ##p_1 \in \gamma## at parameter ##\lambda_1## on the worldline ##\gamma## of an observer, and then reflected at some event ##q## back toward another point ##p_2 \in \gamma## at parameter ##\lambda_2## on the worldline, the event ##p \in \gamma## which is EP-
simultaneous with ##q## is the one at parameter\begin{align*}
\lambda = \dfrac{1}{2}(\lambda_1 + \lambda_2)
\end{align*}Assume, to begin, that the 4-velocity ##\mathbf{u}## of the observer is constant, then\begin{align*}
\overrightarrow{p_1 p} = (\lambda - \lambda_1)\mathbf{u} \\
\overrightarrow{p p_2} = (\lambda_2 - \lambda) \mathbf{u}
\end{align*}Meanwhile the vectors ##\overrightarrow{p_1 q} = \overrightarrow{p_1 p} + \overrightarrow{pq}## and ##\overrightarrow{qp_2} = \overrightarrow{qp} + \overrightarrow{pq_2}## are null (they are the paths of a light ray) and therefore\begin{align*}
\overrightarrow{p_1 q} \cdot \overrightarrow{p_1 q} &= (\overrightarrow{p_1 p} + \overrightarrow{pq}) \cdot (\overrightarrow{p_1 p} + \overrightarrow{pq}) \\
&= (\lambda - \lambda_1)^2 \mathbf{u} \cdot \mathbf{u} + 2 (\lambda - \lambda_1)\mathbf{u} \cdot \overrightarrow{pq} + \overrightarrow{pq} \cdot \overrightarrow{pq} \overset{!}{=} 0 \\ \\
\overrightarrow{qp_2} \cdot \overrightarrow{qp_2} &= (\overrightarrow{qp} + \overrightarrow{pp_2}) \cdot (\overrightarrow{qp} + \overrightarrow{p p_2}) \\
&= \overrightarrow{qp} \cdot \overrightarrow{qp} + 2 (\lambda_2 - \lambda) \mathbf{u} \cdot \overrightarrow{qp} + (\lambda_2 - \lambda)^2 \mathbf{u} \cdot \mathbf{u} \overset{!}{=} 0
\end{align*}Assuming ##\mathbf{u}## to be normalised as ##\mathbf{u} \cdot \mathbf{u} = -1##, and re-writing ##\overrightarrow{qp} = - \overrightarrow{pq}##, this becomes\begin{align*}
-(\lambda - \lambda_1)^2 + 2 (\lambda - \lambda_1)\mathbf{u} \cdot \overrightarrow{pq} + \overrightarrow{pq} \cdot \overrightarrow{pq} = 0 \\
\overrightarrow{pq} \cdot \overrightarrow{pq} - 2 (\lambda_2 - \lambda) \mathbf{u} \cdot \overrightarrow{pq} - (\lambda_2 - \lambda)^2 = 0
\end{align*}subtracting:\begin{align*}
(\lambda_2 - \lambda)^2 - (\lambda - \lambda_1)^2 - 2 ( \lambda_1 + \lambda_2)\mathbf{u} \cdot \overrightarrow{pq} &= 0 \\\implies (\lambda_2 + \lambda_1)(\lambda_1 + \lambda_2 - 2\lambda) - 2( \lambda_1 + \lambda_2)\mathbf{u} \cdot \overrightarrow{pq} &= 0
\end{align*}Therefore ##\lambda = \dfrac{1}{2}(\lambda_1 + \lambda_2) \iff \mathbf{u} \cdot \overrightarrow{pq} = 0##. This is the equation of a hyperplane ##\Pi \subseteq \mathbf{R}^4## with normal ##\mathbf{u}##, and is referred to as a surface of simultaneity.
If the 4-velocity is not constant along the worldline, then the above reasoning still holds providing the points ##q## are sufficiently "close" to the worldline [i.e. small compared to the curvature ##1/(\mathbf{a} \cdot \mathbf{a})##, where ##\mathbf{a} = \dfrac{d\mathbf{u}}{d\lambda}##], so that the curvature of the worldline can be neglected.