In above Figure-01 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{align}
\!\!\!\!\!\!\!\!\!\mathbf{u}_1 \boldsymbol{=} \left(u_{1x},u_{1y},u_{1z}\right) & \boldsymbol{=}\left(u_1 n_{1x},u_1 n_{1y},u_1 n_{1z}\right) \boldsymbol{=} u_1 \mathbf{n}_1\,, \qquad u_1 \in \left(-c,0\right)\cup\left(0,c\right)
\tag{01a}\label{01a}\\
\Vert \mathbf{n}_1 \Vert^2 & \boldsymbol{=} n^2_{1x}\boldsymbol{+}n^2_{1y} \boldsymbol{+} n^2_{1z} \boldsymbol{=}1
\tag{01b}\label{01b}
\end{align}
The Lorentz transformation $\:\mathrm S \longrightarrow \mathrm S'\:$ is
\begin{align}
\mathrm d\mathbf{r'} & \boldsymbol{=} \mathrm d\mathbf{r}\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathrm d\mathbf{r})\mathbf{n}_1\boldsymbol{-}\gamma_1 \mathbf{u}_1\mathrm d t
\tag{02a}\label{02a}\\
\mathrm d t' & \boldsymbol{=} \gamma_1\left(\mathrm d t\boldsymbol{-}\dfrac{\mathbf{u}_1\boldsymbol{\cdot} \mathrm d\mathbf{r}}{c^{2}}\right)
\tag{02b}\label{02b}\\
\gamma_1 & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2_1}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{02c}\label{02c}
\end{align}
while its inverse $\:\mathrm S' \boldsymbol{\longrightarrow} \mathrm S\:$ is
\begin{align}
\mathrm d\mathbf{r} & \boldsymbol{=} \mathrm d\mathbf{r'}\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathrm d\mathbf{r'})\mathbf{n}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1\mathrm d t'
\tag{03a}\label{03a}\\
\mathrm d t & \boldsymbol{=} \gamma_1\left(\mathrm d t'\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot} \mathrm d\mathbf{r'}}{c^{2}}\right)
\tag{03b}\label{03b}
\end{align}
Now, let a point particle $\:\mathrm P\:$ moving with velocity $\:\mathbf{u}_2\:$ with respect to system $\:\mathrm S'\:$ where
\begin{align}
\!\!\!\!\!\!\!\!\!\mathbf{u}_2 \boldsymbol{=}\dfrac{\mathrm d\mathbf{r'}}{\mathrm d t'} \boldsymbol{=}\left(u_{2x'},u_{2y'},u_{2z'}\right) & \boldsymbol{=} \left(u_2 n_{2x'},u_2 n_{2y'},u_2 n_{1z'}\right) \boldsymbol{=} u_2 \mathbf{n}_2\,, \quad u_2 \in \left(-c,c\right)
\tag{04a}\label{04a}\\
\Vert \mathbf{n}_2 \Vert^2 & = n^2_{2x'}+n^2_{2y'} + n^2_{2z'} = 1
\tag{04b}\label{04b}\\
\gamma_2 & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2_2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{04c}\label{04c}
\end{align}
In order to find its velocity $\:\mathbf{u}\:$ with respect to system $\:\mathrm S\:$ where
\begin{align}
\mathbf{u} \boldsymbol{=}\dfrac{\mathrm d\mathbf{r}}{\mathrm d t} \boldsymbol{=}\left(u_{x},u_{y},u_{z}\right) & \boldsymbol{=} \left(u n_{x},u n_{y},u n_{z}\right) \boldsymbol{=} u \mathbf{n}\,, \qquad u \in \left(-c,c\right)
\tag{05a}\label{05a}\\
\Vert \mathbf{n} \Vert^2 & = n^2_{x}+n^2_{y} + n^2_{z} = 1
\tag{05b}\label{05b}\\
\gamma & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{05c}\label{05c}
\end{align}
we divide equations \eqref{03a}, \eqref{03b} side by side and have
\begin{equation}
\mathbf{u} \boldsymbol{=}\dfrac{\mathbf{u}_2\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathbf{u}_2)\mathbf{n}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1}{ \gamma_1\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^{2}}\right)}
\tag{06}\label{06}
\end{equation}
Replacing $\:\mathbf{n}_1\boldsymbol{\longrightarrow} \mathbf{u}_1/u_1\:$
\begin{equation}
\boxed{\:\:\mathbf{u} \boldsymbol{=}\dfrac{\mathbf{u}_2\boldsymbol{+}\dfrac{\gamma^2_{1}\left(\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2\right)}{c^2 \left(\gamma_{1}\boldsymbol{+}1\right)}\mathbf{u}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1}{ \gamma_1\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^{2}}\right)}\:\:}
\tag{07}\label{07}
\end{equation}
Above equation, beyond to be the transformation law for 3-velocities, is the law of relativistic addition of 3-velocities, more exactly it's the relativistic sum of $\:\mathbf{u}_1,\mathbf{u}_2$.
Now, between the $\gamma-$factors $\gamma,\gamma_{1},\gamma_{2}\:$ the following equation is valid
\begin{equation}
\boxed{\:\:
\gamma \boldsymbol{=}\gamma_{1}\gamma_{2}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{08}\label{08}
\end{equation}
This relation is proved as follows :
Let $\:\mathrm S^{\mathrm P}\:$ the rest system of the particle $\:\mathrm P$. In this system $\:\mathrm S^{\mathrm P}\:$ the time is the proper one $\:\tau$. The rest system $\:\mathrm S^{\mathrm P}\:$ is moving with velocity $\:\mathbf{u}_2\:$ with respect to system $\:\mathrm S'\:$ so according to the Lorentz transformation between these systems we have
\begin{equation}
\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\gamma_2
\tag{09}\label{09}
\end{equation}
On the same step, since the rest system $\:\mathrm S^{\mathrm P}\:$ is moving with velocity $\:\mathbf u\:$ with respect to system $\:\mathrm S\:$ we have
\begin{equation}
\dfrac{\mathrm dt\hphantom{'}}{\mathrm d\tau}\boldsymbol{=}\gamma
\tag{10}\label{10}
\end{equation}
On the other hand from the Lorentz transformation between the systems $\:\mathrm S\:$ and $\:\mathrm S'\:$ we have, see \eqref{03b}
\begin{equation}
\dfrac{\mathrm dt\hphantom{'}}{\mathrm dt'}\boldsymbol{=}\gamma_{1}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right)
\tag{11}\label{11}
\end{equation}
From equations \eqref{09},\eqref{10} and \eqref{11} the relation \eqref{08} is proved, that is
\begin{equation}
\gamma\boldsymbol{=}\dfrac{\mathrm dt\hphantom{'}}{\mathrm d\tau}\boldsymbol{=}\dfrac{\mathrm dt\hphantom{'}}{\mathrm dt'}\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\gamma_{1}\gamma_{2}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right)
\tag{12}\label{12}
\end{equation}
In terms of the rapidities $\:\zeta_1,\zeta_2,\zeta\:$ where
\begin{equation}
\tanh\zeta_1\boldsymbol{=}\dfrac{u_1}{c}\,,\quad \tanh\zeta_2\boldsymbol{=}\dfrac{u_2}{c}\,,\quad\tanh\zeta\boldsymbol{=}\dfrac{u}{c}
\tag{13}\label{13}
\end{equation}
equation \eqref{08} is rewritten as
\begin{equation}
\boxed{\:\:\cosh\zeta\boldsymbol{=}\cosh\zeta_1\cosh\zeta_2\boldsymbol{+}\underbrace{\left(\mathbf{n}_1\boldsymbol{\cdot}\mathbf{n}_2\right)}_{\cos\omega}\sinh\zeta_1\sinh\zeta_2\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{14}\label{14}
\end{equation}
where $\:\omega \in [0,\pi]\:$ the angle between the unit vectors $\:\mathbf{n}_1\:$ and $\:\mathbf{n}_2$, see Figure-01.
In case of parallel $\:\mathbf{n}_1,\mathbf{n}_2\:$ we have
\begin{equation}
\zeta\boldsymbol{=}
\left.
\begin{cases}
\zeta_1\boldsymbol{+}\zeta_2 & \text{if}\:\:\: \omega=0\\
\zeta_1\boldsymbol{-}\zeta_2 & \text{if}\:\:\: \omega=\pi
\end{cases}\right\}
\tag{15}\label{15}
\end{equation}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
Now, we'll make a correlation with the composition of two rotations in space.
As we'll see in the following (Figure-04), Figure-02 is used for the geometric representation (construction) of the composition of two rotations in space. In this Figure two planes $\:\rm p_1, p_2\:$ at an angle $\:\omega\:$ intersect on line $\:\epsilon$. Two lines $\:\epsilon_1,\epsilon_2\:$ on planes $\:\rm p_1, p_2\:$ respectively are passing through a common point on line $\:\epsilon\:$ at angles $\:\phi_1,\phi_2\:$ with respect to this line. By elementary trigonometry we have \begin{equation}
\cos\phi \boldsymbol{=} \cos\phi_1 \cos\phi_2\boldsymbol{-}\cos\omega \sin\phi_1 \sin\phi_2
\tag{16}\label{16}
\end{equation}
If in above equation we replace the real angles $\:\phi_1,\phi_2,\phi\:$ with imaginary ones
\begin{equation}
\phi_1\boldsymbol{=}\mathrm{i}\,\zeta_1\,,\quad \phi_2\boldsymbol{=}\mathrm{i}\,\zeta_2\,,\quad \phi\boldsymbol{=}\mathrm{i}\,\zeta
\tag{17}\label{17}
\end{equation}
that is
\begin{equation}
\cos\left(\mathrm{i}\,\zeta\right) \boldsymbol{=} \cos\left(\mathrm{i}\,\zeta_1\right) \cos\left(\mathrm{i}\,\zeta_2\right)\boldsymbol{-}\cos\omega \sin\left(\mathrm{i}\,\zeta_1\right) \sin\left(\mathrm{i}\,\zeta_2\right)
\tag{18}\label{18}
\end{equation}
then we'll meet equation \eqref{14} again
\begin{equation}
\cosh\zeta\boldsymbol{=}\cosh\zeta_1\cosh\zeta_2\boldsymbol{+}\cos\omega\sinh\zeta_1\sinh\zeta_2
\tag{19}\label{19}
\end{equation}
since
\begin{equation}
\cos\left(\mathrm{i}\,\rho\right)\boldsymbol{=}\cosh\rho\,,\quad \sin\left(\mathrm{i}\,\rho\right)\boldsymbol{=}\mathrm{i}\,\sinh\rho \qquad \rho \in \mathbb{R}
\tag{20}\label{20}
\end{equation}
and formally we have Figure-03.
