**Insert diagram of coordinate systems.
$$ \vec r=r_x\hat{e}_x+r_y\hat{e}_y $$
$$ \vec v=\dot r_x\hat{e}_x+\dot r_y\hat{e}_y $$
$$ \vec a=\ddot r_x\hat{e}_x+\ddot r_y\hat{e}_y $$
Polar unit vectors are defined by a counter-clockwise (negative) rotation of the cartesian vectors. The radial direction points from the origin of the coordinate system to a point of interest while the angular direction points 90 degrees (right-handed) from the radial direction. Consider how a velocity expressed in polar coordinates at $\theta$ would need to be rotated by an additional positive $\theta$ in order to be correctly expressed in cartesian coordinates. Thus, the conversion in the other direction, from cartesian to polar coordinates, is defined by a negative rotation.
$$ \begin{bmatrix} \hat{e_r}\\\hat{e_\theta} \end{bmatrix}= \begin{bmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta) \end{bmatrix} \begin{bmatrix} \hat{e_x}\\\hat{e_y} \end{bmatrix} $$
$$ \vec r = r\hat e_r $$
$$ \vec v =\dot r\hat{e}r+r\dot\theta\hat{e}\theta $$
$$ \vec a =(\ddot r-r\dot\theta^2)\hat{e}r +(r\ddot\theta+2\dot r\dot\theta)\hat{e}\theta $$
Path unit vectors are defined again by an angle of rotation, $\beta$, so the same relationship with the cartesian unit vectors applies as did with polar coordinates.
$$ \begin{bmatrix} \hat{e_r}\\\hat{e_n} \end{bmatrix}= \begin{bmatrix} \cos(\beta)&\sin(\beta)\\ -\sin(\beta)&\cos(\beta) \end{bmatrix} \begin{bmatrix} \hat{e_x}\\\hat{e_y} \end{bmatrix} $$
The tangential direction is tangent to the path in the direction of the velocity of a particle; the normal direction is 90 degrees from the tangential direction pointing to the inside of the curvature of the path. Depending on curvature of the path and the direction of the velocity, the normal direction may be right-handed or left-handed with respect to the tangential direction. The previous rotation matrix only applies if the path coordinates are right-handed. If the path coordinates are left-handed, then the following rotation matrix is used to transform cartesian coordinates to path coordinates:
$$ \begin{bmatrix} \hat{e_r}\\\hat{e_n} \end{bmatrix}= \begin{bmatrix} \cos(\beta)&\sin(\beta)\\ \sin(\beta)&-\cos(\beta) \end{bmatrix} \begin{bmatrix} \hat{e_x}\\\hat{e_y} \end{bmatrix} $$
Because the path is prescribed, the position vector is not typically expressed, but rather the speed along that path. This is given by the radius of curvature, $\rho$, and the angle of the path, $\beta$.
$$ \vec v=\rho\dot\beta\hat e_t $$
$$ \rho=\frac{(1+(y')^2)^{\frac 32}}{|y''|} $$
$$ y'=\frac{dy}{dx}=\frac{v_y}{v_x}=\tan\beta $$
The application of the identity $v=\rho\dot\beta$, from the above, gives various expressions for acceleration.
$$ \vec a=a_t\hat e_t+a_n\hat e_n $$
$$ a_t=\dot v=\rho\ddot\beta $$
$$ a_n=v\dot\beta=v^2/\rho=\rho\dot\beta^2 $$
It is crucial to realize that the magnitudes of velocity and acceleration are equivalent no matter which coordinate system they are expressed in.
Some geometrical problem solving may be assisted by employing vector diagrams.