Connecting 4-velocity to equation for adding velocities Is there a way to connect 4-velocity to equations for adding speeds? I know 4-velocity $U^\mu$ is derived like this: 

\begin{equation}
\begin{split}
P^\mu &= m U^\mu \Longrightarrow 
U^\mu = P^\mu \frac{1}{m} =
\begin{bmatrix}
p_x\\
p_y\\
p_z\\
\frac{W}{c}
\end{bmatrix}
\frac{1}{m} =
\begin{bmatrix}
\frac{mv_x \gamma(v)}{m}\\
\frac{mv_y \gamma(v)}{m}\\
\frac{mv_z \gamma(v)}{m}\\
\frac{mc^2 \gamma(v)}{c m}\\
\end{bmatrix}
=
\begin{bmatrix}
v_x \gamma(v)\\
v_y \gamma(v)\\
v_z \gamma(v)\\
c \gamma(v)\\
\end{bmatrix}
\end{split}
\end{equation}
Or like this:
\begin{equation}
\begin{split}
U^\mu = \frac{d X^\mu}{d \tau} = \frac{d}{dt}X^\mu \gamma(v)= \frac{d}{dt}
\begin{bmatrix}
d x\\ d y\\ d z \\ c d t
\end{bmatrix}
\gamma(v) =
\begin{bmatrix}
v_x\gamma(v)\\
v_y\gamma(v)\\
v_z\gamma(v)\\
c \gamma(v)
\end{bmatrix}
\end{split}
\end{equation}

And i know how to derive Lorentz transformations for velocities $\perp$ to relative velocity $u$ (which is in direction of $x$, $x'$ axis) and $\parallel$ to $u$. It goes like this:

a.) $\parallel$ to $u$:
\begin{equation}
\begin{split}
v_y' &= \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\
&= \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)}\\
&\boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}}
\end{split}
\end{equation}
b.) $\perp$ to $u$:
\begin{equation}
\begin{split}
v_x' &= \frac{dx'}{dt'}=\frac{\gamma (d x - u d t)}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ 
&= \frac{d x - u d t}{d t - d x \frac{u}{c^2}} = \frac{\frac{d x}{d t} - u \frac{d t}{d t}}{\frac{d t}{d t} - \frac{d x}{d t} \frac{u}{c^2}}\\
&\boxed{v_x' = \frac{v_x - u}{1- v_x \frac{u}{c^2}}}
\end{split}
\end{equation}

QUESTION: 
Where is the connection among 4-velocity and equations derived under a.) and b.)? How can i show the connection?
 A: Yes, from the transformation law for the four-velocity, we can explicitly derive the transformations of three-velocities parallel to and perpendicular to a given boost.
First, we need to talk about what the transformation law is for the four-velocity with respect to a boost.  Let the four-velocity be $U = (U^t, U^x, U^y, U^z)$.  Let's boost this along the x-direction by a speed $\mu c$.  Like any four-vector, the four-velocity transforms under Lorentz transformations like so:
$$\begin{align*}
{U'}^t &= W(U^t - \mu U^x) \\
{U'}^x &= W(U^x - \mu U^t) \\
{U'}^y &= U^y \\
{U'}^z &= U^z\end{align*}$$
where $W = 1/\sqrt{1-\mu^2}$ is the Lorentz factor of the boost.
Now, break down the original four-vector $U$ as $U= \gamma c(1, \beta^x, \beta^y, \beta^z)$.  We can find the components of $U'$ as
$$\begin{align*}
{U'}^t &= W\gamma c(1 - \mu \beta^x) \\
{U'}^x &= W \gamma c (\beta^x - \mu) \\
{U'}^y &= \gamma c \beta^y \\
{U'}^x &= \gamma c \beta^z\end{align*}$$
You can then find the components of three-velocity in the primed frame by taking ${U'}^i/{U'}^t$.
$$\begin{align*}
\frac{{v'}^x}{c} &= \frac{{U'}^x}{{U'}^t} = \frac{\beta^x - \mu}{1 - \mu \beta^x} \\
\frac{{v'}^y}{c} &= \frac{{U'}^y}{{U'}^t} = \frac{\beta^y}{W(1-\mu \beta^x)} \\
\frac{{v'}^z}{c} &= \frac{{U'}^z}{{U'}^t} = \frac{\beta^z}{W(1-\mu \beta^x)}
\end{align*}$$
These are algebraically the same as the formulas you posted in (a) and (b).  To me, this is much simpler than churning through velocity-addition.  The transformation laws of four-vectors are simple to learn, and the amount of manipulation needed to find the new three-velocity is minimal.
*Note: your terminology on parallel vs. perpendicular to the boost seems confused.  Nevertheless, I believe my results here capture what you intended.  The boost velocity is in the direction of the x-axis.
