Intuitively understanding the regular curvature $\frac{\|\det(r'(t),r''(t))\|}{\|r'(t)\|^3}$ 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)
 A: 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.
