A best definition of proper acceleration In this link of Wikipedia https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity) there is a definition of proper acceleration:

The proper acceleration to a particle is defined as the acceleration that a particle "feels" while accelerating from one inertial reference system to another.
$$\alpha=\frac{1}{(1-u^2/c^2)^{3/2}}\frac{du}{dt}.$$

May I ask, please, if there is a better alternative definition to this or possibly some documentation in English clearer than that of Wikipedia?
 A: Restricting consideration to $(1+1)$-Minkowski spacetime,
I think the best definition of the proper acceleration of a worldline ("worldline curvature")
is essentially the derivative of its rapidity $\theta$ with respect to its proper time $\tau$
\begin{align}
\alpha\equiv c \frac{d\theta}{d\tau},
\end{align}
where $u=c\tanh\theta$ and $\gamma=\displaystyle\frac{1}{\sqrt{1-(u/c)^2}}=\cosh\theta$
and the 4-vector $d\hat\tau = \cosh\theta\ d \hat t + \sinh\theta\ d \hat x$.
To connect this with your expression,
write
\begin{align}
\alpha\equiv c \frac{d\theta}{d\tau}=c\frac{d\theta}{dt} \frac{dt}{d\tau}=c\frac{d\theta}{dt} \cosh\theta,
\end{align}
then
note that: since $u=c\tanh\theta$,
the coordinate acceleration is
\begin{align}
a=\frac{du}{dt}
&=\frac{d}{dt}( c\tanh\theta )\\
&=\frac{d\theta}{dt}\frac{d}{d\theta}(c \tanh\theta )\\
&=\frac{d\theta}{dt}\frac{c}{\cosh^2\theta}\\
\end{align}
So,
we have
\begin{align}
\alpha\equiv c \frac{d\theta}{d\tau}
&=c\frac{d\theta}{dt} \frac{dt}{d\tau}\\
&=\left( a \cosh^2\theta \right) \cosh\theta\\
&= a \cosh^3\theta\\ 
&= a \left( \frac{1}{\sqrt{1-(u/c)^2}}\right)^3
\end{align}
Update2
The above expression for $\alpha$ (and what was given in the OP)
is the spacetime-analogue of the "curvature of a plane curve"
in Euclidean geometry:
https://en.wikipedia.org/wiki/Curvature#Graph_of_a_function
writes $k=\frac{y''}{\left(1+y'^2\right)^{3/2}}$
which we rewrite as
$$k=y''\left( \frac{1}{\sqrt{1+(y')^2}}\right)^3$$

Update
The rapidity $\theta$ is Minkowski-analogue of the "angle between 4-velocities".
It's useful to compare the

rapidity and the hyperbolic trig functions in Minkowski-spacetime-geometry

and
angle and the circular trig functions in Euclidean-geometry.
(In my opinion, if you expect students to be comfortable with functions like $(1-(u/c)^2)^{-3/2}$, they could be comfortable with hyperbolic trig functions (assuming they are familiar with the circular functions and trigonometry)).
Below, I have attached a Minkowski-spacetime diagram  [time runs upwards by convention].
Imagine the tips of all possible 4-velocities from a given event...
their tips trace out a hyperbola [with spacelike tangents].

The rapidity is the spacelike-arclength along a hyperbola [with spacelike-tangents], divided by the radius.
(The angle is the arc-length on a circle, divided by the radius of the circle $\theta_E=S_{circ}/R_{circ}$.)
Velocity $u=c\tanh\theta$. ("Slope" is $m=\tan\theta_E$). (Note while the slopes of a line may be equal, their angles in their respective geometries are not... however, the analogous tangent-functions of those angles are equal.)
The ratio of the adjacent side to the hypotenuse is $\cosh\theta=\frac{1}{\sqrt{1-\tanh^2\theta}}=\frac{1}{\sqrt{1-(u/c)^2}}=\gamma$, which is the time-dilation factor. (The ratio of the adjacent side to the hypotenuse is $\cos\theta_E=\frac{1}{\sqrt{1+\tan^2\theta_E}}$.)
Note:
$\cosh^2\theta(1-\tanh^2\theta) \equiv \cosh^2\theta-\sinh^2\theta \equiv 1$

($\cos^2\theta_E(1+\tan^2\theta_E) \equiv \cos^2\theta_E+\sin^2\theta_E \equiv 1$)
$(u_{rel}/c)=\tanh(\theta_2-\theta_1)=\frac{\tanh\theta_2-\tanh\theta_1}{1-\tanh\theta_2\tanh\theta_1}=\frac{(u_2/c)-(u_1/c)}{1-u_2 u_1/c^2}$

($m_{rel}=\tan(\theta_{E2}-\theta_{E1})=\frac{\tan\theta_{E2}-\tan\theta_{E1}}{1+\tan\theta_{E2}\tan\theta_{E1}}=\frac{m_2-m_1}{1+m_2 m_1}$)
Note: a unit-timelike vector has the form
\begin{align}
d\hat\tau
&=\cosh\theta\ d\hat t+\sinh\theta\ d\hat x\\
&=\cosh\theta(  d\hat t+\tanh\theta\ d\hat x)\\
&=\gamma(  d\hat t+ u \ d\hat x)\\
&=\gamma\  d\hat t+ \gamma u \ d\hat x,
\end{align}
where $d\hat t\cdot d\hat t=1$, $\ d\hat x\cdot d\hat x=-1$,
and $\ d\hat t\cdot d\hat x=0$. [Verify that $d\hat\tau \cdot d\hat\tau =1$.]

(A unit-vector in Euclidean space has the form
$$d\hat s=\cos\theta_E\ d\hat x + \sin\theta_E\ d\hat y,$$
which could be written in terms of slopes instead of angles
$$d\hat s=\frac{1}{\sqrt{1+m^2}}\ d\hat x + \frac{m}{\sqrt{1+m^2}}\ d\hat y.$$
In Euclidean space,
$d\hat x\cdot d\hat x=1$, $\ d\hat y\cdot d\hat y=1$,
and $\ d\hat x\cdot d\hat y=0$. [Verify that $d\hat s \cdot d\hat s =1$.])


Next,
I have attached a spacetime diagram of a uniformly-accelerated worldline.
I have drawn the unit-tangent vector (the 4-velocity) to that worldline at equal intervals of the accelerated-observer's proper time $\tau$. At those events, I have also drawn the 4-velocity of the observer at rest in this diagram.
For a uniformly-accelerated observer, the change in rapidity $\Delta\theta$ is proportional to the change in that observer's proper-time $\Delta\tau$ .
(For a circle, the change in the angle of the tangent is proportional to the change in arc-length along the circle. See this summarized [with equations analogous to my equations above] at http://mathworld.wolfram.com/Curvature.html ).

A: Let's say there is an inertial frame S close to an accelerated frame S'. S' velocity is not constant, but at every moment, we can assume S velocity to be exactly same as S' velocity. Now if S measures S' acceleration, what he will acquire would be proper acceleration. Or if you stay in a rocket that its engines provides constant force (i.e you will feel a constant weight in rocket), then what you measure for acceleration is constant proper acceleration.
You can check Explorations in Mathematical Physics from DonKoks for more details. It uses simple notions without tensor and such in accelerated frames section.
A: As far as the definition of (the magnitude of) proper acceleration of a participant $A$ in event $\varepsilon_{A \Xi}$ simply is (proportional to) the curvature, at this event, of the world line $\mathcal W_A$ traced by $A$ (i.e. of the ordered set of events in which $A$ took part) it can be expressed directly in terms of spacetime intervals between pairs of these events alone.
(Arguably, that's simpler than referring to any description of $A$ wrt. some suitable reference frame as intermediary.)
Denote as $\mathcal P_A[ \, \Xi \, ] := \{ \, \varepsilon_{A \Psi} \in \mathcal W_A : \varepsilon_{A \Psi} \ll \varepsilon_{A \Xi} \, \}$ the subset of $A$'s world line consisting of events which chronologically precede event $\varepsilon_{A \Xi}$;
and as $\mathcal F_A[ \, \Xi \, ] := \{ \, \varepsilon_{A \Phi} \in \mathcal W_A : \varepsilon_{A \Xi} \ll \varepsilon_{A \Phi} \, \}$ the subset of $A$'s world line consisting of events which chronologically follow event $\varepsilon_{A \Xi}$.
Then the magnitude of $A$'s proper acceleration at event $\varepsilon_{A \Xi}$ can be expressed as
$$\normalsize | a_A[ \, \Xi \, ] | := c \left( \! \! \overset{{\huge \text{lim}}}{\small \overset{}{\begin{matrix}\varepsilon_{A \Psi} \in \mathcal P_A[ \, \Xi \, ], \cr \varepsilon_{A \Phi} \in \mathcal F_A[ \, \Xi \, ]  : \cr s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \rightarrow 0 \end{matrix}}} \! \! \left[ \, \frac{
\begin{vmatrix} 0 & 1 & 1 & 1 \cr 1 & 0 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] & s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] \cr
1 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] & 0 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \cr 1 & s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] & 0
\end{vmatrix}}{-{\, \, \, \overset{}{| \,  s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] \,  s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] \,  s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \, | \, \, \, }}} \, \right] \right)^{\! \! \! \left(\frac{1}{2}\right)}$$
where the curvature is the inverse of the radius of curvature, the formula for timelike triangles corresponds to the suitably signed expression for euclidean triangles with a Cayley-Menger determinant in the numerator, and with proportionality factor $c$ (signal front speed).
Further, since all spacetime intervals in this expression are timelike, the generalization to non-flat regions is readily obtained by replacing $s^2$ by the square of Lorentzian distance, $\ell^2$.
(The direction of any non-zero proper acceleration of $A$ in event $\varepsilon_{A \Xi}$ can be characterized in relation to the worldline of a non-accelerating participant who was momentarily co-moving with $A$ in event $\varepsilon_{A \Xi}$.)
A: As far as I know, the definition of the proper acceleration was the derivative of the proper velocity with respect to proper time, $$\vec{\alpha}=\frac{\mathrm{d} \vec{w}}{\mathrm{d} \tau}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} \tau^2}.$$ It is the spatial part of the four acceleration, and in Wikipedia in English it says that the equation that you cite is valid iff the proper acceleration is directed parallel to the line of motion. The relation between proper acceleration and ordinary acceleration is instead $$\vec{\alpha}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \gamma^2+\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \frac{\mathrm{d} \gamma}{\mathrm{d} \tau}$$ by the chain rule. You can identify $\vec{a}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2}$ as the ordinary acceleration.
