Lorentz transformation of the four velocity Can we derive the velocity addition rule by directly transforming the four velocity?
 A: Theory-wise, the easiest way to use this is to use $w=ct$ and then your Lorentz matrix in 1+1 dimensions can be written as $$\begin{bmatrix}w'\\x'\end{bmatrix} = \begin{bmatrix}\cosh\alpha & -\sinh\alpha\\-\sinh\alpha&\cosh\alpha\end{bmatrix}\begin{bmatrix}w\\x\end{bmatrix}.$$
This is because the hyperbolic cosine and sine obey the relation $\cosh^2\alpha - \sinh^2\alpha = 1,$ and therefore if you naturally just choose $\cosh\alpha = 1/\sqrt{1 - \beta^2}$ you naturally find $\sinh^2\alpha = \beta^2/(1 - \beta^2),$ so this is a valid way to write the usual Lorentz Matrix $[\gamma, -\gamma\beta; -\gamma\beta, \gamma].$
Since a 4-velocity in its rest frame is $c~[1; 0]$ it is not hard to see straight from this that a 4-velocity in any other reference frame is $c~[\cosh r; \sinh r]$and therefore that $r = \tanh^{-1}(v/c)$ where $v$ is the ordinary velocity and $r$ is this new quantity called the rapidity. Plugging that in, we can find that it's Lorentz transform is $$\begin{bmatrix}\cosh\alpha & -\sinh\alpha\\-\sinh\alpha&\cosh\alpha\end{bmatrix}\begin{bmatrix}\cosh r\\\sinh r\end{bmatrix} = \begin{bmatrix}\cosh r~\cosh\alpha - \sinh r~\cosh \alpha\\\sinh r~\cosh\alpha-\cosh r~\sinh\alpha\end{bmatrix} = \begin{bmatrix}\cosh (r -
 \alpha)\\\sinh(r-\alpha)\end{bmatrix}.$$So if you transform to its rest frame for example you choose $\alpha=r$ and then the 4-velocity takes this $[c; 0]$ form directly, but rapidities add linearly in one dimension.
Of course for this particular question all of that theory-which-makes-things-simple can be a little bit overkill. If you do not want to use the hyperbolic sines and cosines, you can write this out in terms of the more conventional components as $$\gamma_2\begin{bmatrix}1\\\beta_2\end{bmatrix}= \gamma_0~\gamma_1~\begin{bmatrix}1&-\beta_0\\-\beta_0&1\end{bmatrix}\begin{bmatrix}1\\\beta_1\end{bmatrix}.$$
Notice that the noisy prefactors $\gamma_{0,1,2}$ do not need to enter our consciousness at all because the ratio of the second component to the first will be $\gamma_2 \beta_2 / \gamma_2,$ and they will just cancel out on either side So all you're left with when you take this ratio is,$$\beta_2 = \frac{\beta_1 - \beta_0}{1 - \beta_0\beta_1}.$$
A: 
Instead of the well-known 1-space-dimensional Lorentz Transformation the more general 3-space-dimensional one for the configuration of above Figure (see its 3D version in the end) and with finite variables is(1)
\begin{align}                 
 \mathbf{x}^{\boldsymbol{\prime}} & =  \mathbf{x}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot}  \mathbf{x})\mathbf{n}+\gamma_{v} \mathbf{v} t 
\tag{01a}\\
 t^{\boldsymbol{\prime}}& =  \gamma_{v}\left( t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right)
\tag{01b}      
\end{align}
where
\begin{equation}
  \mathbf{n}\equiv \dfrac{\mathbf{v}}{\| \mathbf{v}\| }=\dfrac{\mathbf{v}}{v}\, , \qquad \gamma_{v} = \gamma\left(v\right)\equiv \left(1-\dfrac{v^{2}}{c^{2}}\right)^{-\frac12}
\tag{02}
\end{equation}
and because of linearity with differentials
\begin{align}                 
   \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}} & =   \mathrm{d}\mathbf{x}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot}  \mathrm{d} \mathbf{x})\mathbf{n}+\gamma_{v} \mathbf{v} \mathrm{d} t
\tag{03a}\\
 \mathrm{d} t ^{\boldsymbol{\prime}}& =  \gamma_{v}\left( \mathrm{d} t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}}{c^{2}}\right)
\tag{03b}      
\end{align}
Dividing (03a) by (03b) we have
\begin{equation}
  \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma_{v}-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma_{v} \mathbf{v}\vphantom{\dfrac12}}{\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot}  \mathbf{u}}{c^{2}}\Biggr)}
\tag{04}
\end{equation}
Equation (04) is a more general "rule" for the addition of velocities $\mathbf{v},\mathbf{u}$. The 1-dimensional version is the well-known equation 
\begin{equation}
 u^{\boldsymbol{\prime}} = \dfrac{u+v }{1+\dfrac{v\, u}{c^{2}}\vphantom{\dfrac12}}
\tag{05}
\end{equation}
Now, the 4-velocity
\begin{equation}
\mathbf{U} = \left(\gamma_{u}\mathbf{u} ,\gamma_{u} c\right)\, , \qquad \gamma_{u} = \gamma\left(u\right)\equiv \left(1-\dfrac{u^{2}}{c^{2}}\right)^{-\frac12} 
\tag{06}
\end{equation}
as could be proved(2), is a 4-vector : it is Lorentz-transformed as the space-time position vector
\begin{equation}
\mathbf{X} = \left(\mathbf{x} ,c\,t\right)
\tag{07}
\end{equation}

I don't understand why do you want to reach (04) from the fact that (06) is a 4-vector when inversely this property is a consequence of (04).


(1)
The Lorentz Transformation (01) could be expressed in block matrix form as 
\begin{equation}
\mathbf{X'}
=
\begin{bmatrix}
\mathbf{x'}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
c\,t'\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} 
\end{bmatrix}
=
\begin{bmatrix}
         \mathrm{I}+(\gamma_{v}-1)\mathbf{n}\mathbf{n}^{\top} & \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}\\
        &\\
         \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}^{\top} &\hspace{5mm}\gamma\\                         
   \end{bmatrix}
\begin{bmatrix}   
\mathbf{x}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
c\,t\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} 
\end{bmatrix} 
=\Lambda \mathbf{X}
\tag{fn-01}
\end{equation}
where $\:\mathbf{n}\:$ a $\:3\times 1\:$ unit column vector and $\: \mathbf{n}^{\top} \:$ its transposed  $\:1\times 3\:$ unit row vector
\begin{equation}
\mathbf{n}=
\begin{bmatrix}
   n_1\\         
   n_2 \\         
   n_3            
\end{bmatrix}
\quad , \quad
\mathbf{n}^{\top}=
\begin{bmatrix}
  n_1&n_2&n_3
\end{bmatrix}
\tag{fn-02}
\end{equation}
and $\:\mathbf{n}\mathbf{n}^{\top}\:$  a linear transformation, the vectorial projection on the direction $\:\mathbf{n}\:$
\begin{equation}
\mathbf{n}\mathbf{n}^{\top}=
\begin{bmatrix}
   n_1\\         
   n_2 \\         
   n_3            
\end{bmatrix}
\begin{bmatrix}
  n_1&n_2&n_3
\end{bmatrix}
=
\begin{bmatrix}
   n_1^{2} & n_1 n_2 & n_1 n_3\\         
   n_2 n_1  & n_2^{2} & n_2 n_3\\          
   n_3 n_1  & n_3 n_2 & n_3^{2}         
\end{bmatrix}
\tag{fn-03}
\end{equation}


(2)
If a particle is moving with velocity $\mathbf{u}$ with respect to system $\mathrm{S}$ then between its proper time $\tau$ and times $t,t'$ we have
\begin{equation}
\dfrac{\mathrm{d}t}{\mathrm{d}\tau}=\gamma_{u}\, , \quad \dfrac{\mathrm{d}t'}{\mathrm{d}\tau}=\gamma_{u'}
\tag{fn-04}
\end{equation}
but from (03b)
\begin{equation}
\dfrac{\mathrm{d}t'}{\mathrm{d}t}=\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot}  \mathbf{u}}{c^{2}}\Biggr)
\tag{fn-05}
\end{equation}
so
\begin{equation}
\gamma_{v} \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot}  \mathbf{u}}{c^{2}}\Biggr)=\dfrac{\gamma_{u'}}{\gamma_{u}}
\tag{fn-06}
\end{equation}
Replacing this in the dominator of the rhs of (04) we note that the quantities $\gamma_{u}\mathbf{u}$ and $\gamma_{u}$ are transformed as  $\mathbf{x}$ and $t$ in equations (01), that is finally
\begin{equation}
\mathbf{U'}
=
\begin{bmatrix}
\gamma_{u'}\mathbf{u'}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
\gamma_{u'}c\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} 
\end{bmatrix}
=
\begin{bmatrix}
         \mathrm{I}+(\gamma_{v}-1)\mathbf{n}\mathbf{n}^{\top} & \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}\\
        &\\
         \dfrac{\gamma_{v} \upsilon}{c}\mathbf{n}^{\top} &\hspace{5mm}\gamma\\                         
   \end{bmatrix}
\begin{bmatrix}   
\gamma_{u}\mathbf{u}\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} \\
\\
\gamma_{u}c\vphantom{\dfrac{\gamma \upsilon}{c}\mathbf{n}} 
\end{bmatrix} 
=\Lambda \mathbf{U}
\tag{fn-07}
\end{equation}



A: Of course you can. You just have to use the definition. Don't get lost in the Minkowski space; velocity is the time derivative of the position. 
If you derive the expression
$x'=\gamma (x-vt)$
you will finally obtain the transformations of velocities; but be careful, because deriving respect to $t$ is not the same as doing so with $t'$. You have to apply the chain rule. The development is done in books, like Alonso-Finn.
Of course this is not the only way to get them. Using the parameters so that 
$v/c=\tanh \zeta$, then the addition formula is directly obtained as the sum of two $\tanh$'s. 
