I wonder why the regular curvature for a parametric arc $r(t)$ is defined by
$$ \frac{\|\det (r'(t) , r''(t))\|}{ \|r'(t)\|^3}.$$
Is it considered as a definition , if yes why it is logical?
How could we explain intuively that definition?
Otherwise, what is a more general curvature definition (I mean for parametric arc)
No it is not an initial definition. If it were the initial definition, then it would sound artificial and non natural.
For a plane curve $\gamma(s)$, $s$ is the length parameter, it is natural to define the curvature as $d\phi/ds$ where $\phi$ is the angle between tangent vector $\dot \gamma$ and the horizontal direction. It is really a natural and reasonable definition.
Now it is easy to prove that $|d\phi/ds|=|\ddot \gamma(s)|$.
On the other hand the term $$(1)\;\;\; \frac{\|\det (r'(t) , r''(t))\|}{ \|r'(t)\|^3}$$ is independednt of parametrizatioin in the following sense:
IF $\alpha(t)$ is a curve and $t=J(s)$ is a diffeomorphism in parameter space and we put $\beta(s)=\alpha(J(s))$ $then we have
$$(1)'\;\; \frac{\|\det (\alpha'(t) , \alpha''(t))\|}{ \|\alpha'(t)\|^3}= \frac{\|\det (\beta'(s) , \beta''(s))\|}{ \|\beta'(t)\|^3}.$$
Now if we compute $(1)$ or $(1)'$ for a unit speed curve $\gamma(s)$ we get $|\ddot \gamma (s)|$, the natural and reasonable definition of curvature.
So it is natural to define the curvature of a space curve $\gamma$ as $|\ddot \gamma(s)|$ provided $\gamma$ is a unit speed parametrization.
On the other hand it can be shown that $$(2)\;\; \frac{\| (r'(t) \times r''(t))\|}{ \|r'(t)\|^3}$$
is independent of parametrization and is equal to $|\ddot \gamma|$.
These materials are mentioned in "Elements of Diff. Geometry" by Andrew Pressley.