Normal Tangential Coordinates 
For normal and tangential coordinates, I seen people say that the unit vectors are changing. But the coordinate axis are changing too. For example, look at $u_t$. (This means u subscript "t")
At first, $u_t$ is at (0,1). But then because the coordinate axis is changing, isn't it at (0,1) for all times? So, I do not understand how $u_t$ is changing. I think I am missing something crucial here and would appreciate any help.
 A: This is an update of my earlier answer, clarified with figures added.
A non-Cartesian coordinate system can be used to specify the position of a particle with respect to the fixed origin of an inertial reference frame.  For example, Figure 1 uses polar coordinates $r$ and $\theta$ to specify the position of a particle at point P with respect to the origin $O$.  The position is specified by the vector $\vec r$ with respect the the origin at $O$. Using a different origin $O^{'}$, the position is described by $\vec r^{'}$.  The unit vectors $\hat n$ and $\hat l$ with respect to $O$ for the polar coordinate system have directions determined by $\vec r$ using origin $O$ and change directions as $\vec r$ changes direction. Similarly, the unit vectors $\hat n^{'}$ and $\hat l^{'}$ with respect to $O^{'}$ have directions determined by $\vec r^{'}$ and change directions as $\vec r^{'}$ changes direction.  Given the total external force $\vec F$ on the particle as a function of time, using Newton's first law the motion of the particle- the path of the particle or the motion of P- can be determined as $\vec r(t)$ or $\vec r^{'}(t)$.
Given the path of the particle in an inertial system as indicated in Figure 2, we can define unit vectors $\hat t$ and $\hat c$ at any point P along the path. $\hat t$ is tangential to the path and points to the direction of positive velocity. $\hat c$ is normal to the path and points toward the center of curvature of the path.  These directions of these unit vectors are not defined with respect to the origin of an inertial system; they are defined at point P dependent on the shape of the path at P.  Using these vectors we can determine the velocity and acceleration at any point P along the path if we know the path. See the MIT video at https://ocw.mit.edu/courses/2-003sc-engineering-dynamics-fall-2011/resources/definition-of-normal-and-tangential-coordinates/.

A: $\def \b {\mathbf}$
assume your curve is a circle.  thus the components of the position vector  are:
\begin{align*}
 &\b R=r\,\begin{bmatrix}
           \cos\left(\frac{s}{r}\right) \\
           \sin\left(\frac{s}{r}\right)\\
         \end{bmatrix}\quad\Rightarrow\\
  &\b t_g=\frac{\partial \b R}{\partial s}=  
 \begin{bmatrix}
           -\sin\left(\frac{s}{r}\right) \\
           \cos\left(\frac{s}{r}\right)\\
         \end{bmatrix}\\
  &\b n= \frac{\partial \b t_g}{\partial s}=
  \begin{bmatrix}
           -\cos\left(\frac{s}{r}\right) \\
           -\sin\left(\frac{s}{r}\right)\\
         \end{bmatrix}\\            
\end{align*}
where r is the radius , s parameter on the path  $~s=s(t)~$ , $~\b t_g~$ is the tangent vector $~\b n~$ is the normal vector, both are function of the parameter s with $~\b t_g\cdot\b n=0$
the velocity vector in inertial system
\begin{align*}
 \b v=\frac{\partial \b R}{\partial s}=\dot{s}\begin{bmatrix}
           -\sin\left(\frac{s}{r}\right) \\
           \cos\left(\frac{s}{r}\right)\\
         \end{bmatrix}
\end{align*}
thus the velocity toward the tangent vector
\begin{align*}
  & v_t=(\b v\cdot \b t_g)=\dot{s}
\end{align*}
and the velocity toward the normal  vector
\begin{align*}
  & v_n=(\b v\cdot \b n)= 0
\end{align*}
