Suppose we have a (four dimensional) Lorentz space; let $T,Z,X$ be instantaneous observers: $X=\g(T+Y)$, $Z=\b(T+vE)$ for $Y,E\in T^\perp$, $\Vert Y\Vert,v < 1$, $\Vert E\Vert=1$ and $\norm Y < 1$. Then the velocity of $X$ with respect to $Z$ is given by definition by
$$
Y^\prime=\frac{X+\la X,Z\ra Z}{-\la X,Z\ra}~.
$$
Let $u$ be the boost $T\to Z$, $E\to W\colon=\b(vT+E)\to E$, $u|\lhull{T,E}^\perp=id$. Then $Y\mapsto-Y^\dprime\colon=-u^{-1}(Y^\prime)$ is a transformation on $B^3\colon=\{\Vert Y\Vert < 1\}$, which extends the Möbius transformation $S^2\rar S^2$.
In this case we have (remember $Y=eE+fF+gG$ and $W=\b(vT+E)$):
$$
-\la X,Z\ra=\b\g(1-ev),
\la X,W\ra=\b\g(-v+e),
\la Z,W\ra=0
$$
and thus
\begin{eqnarray*}
\la Y^\prime,W\ra
&=&\frac{\la X,W\ra}{-\la X,Z\ra}
=\frac{\b\g(-v+e)}{\b\g(1-ev)}\\
\la Y^\prime,F\ra
&=&\frac{\g f}{\b\g(1-ev)},\\
\la Y^\prime,G\ra
&=&\frac{\g g}{\b\g(1-ev)}
\end{eqnarray*}
i.e.
$$
Y^\prime=\frac{e-v}{1-ev}W+\frac{f}{\b(1-ev)}F+\frac{g}{\b(1-ev)}G
$$
Which is exactly the same formula we obtained for light-like vectors $X$.
$$
-Y^\dprime=\frac{v-e}{1-ev}E-\frac{f}{\b(1-ev)}F-\frac{g}{\b(1-ev)}G~.
$$