The Jacobian map is for transforming vectors expressed in terms of one set of coordinate basis vectors into another coordinate system's basis vectors. Positions like $(x,y)$ and $(r,\theta)$ are not expressed in terms of coordinate basis vectors, so it's inappropriate to use the Jacobian to try to convert between them.
Let $e_1, e_2$ be a pair of basis vectors. We can express positions on the 2d plane as $p = x e_1 + y e_2$.
Now, let $f(p) = p' = r e_1 + \theta e_2$. This looks like a change of coordinates, but it's really not--it's an active deformation of the plane into something where $r, \theta$ are "Cartesian" coordinates. This is just an active transformation, however, and fully equivalent to the passive change of coordinates that you're used to.
$f$ is appropriate to move positions to new positions, but it is not appropriate to move, for example, the tangent vector to a curve from one space to another (that is, to express such a tangent vector in terms of the polar coordinate basis vectors). For this, we need the Jacobian map $J_f$.
Example: let $\ell(t) = e_1 \cos t + e_2 \sin t$ be a curve that draws out the unit circle. It's clear that its derivative is the tangent vector $\dot \ell(t) = -e_1 \sin t + e_2 \cos t$. We can't transform this tangent vector using $f$; we must use $J_f$ instead.
(You'll note here I'm moving from Cartesian coordinates to polar, backwards from what you wanted* but the math is basically the same.)
Here's the Jacobian map:
$$\begin{align*}
J_f(e_1) &= \frac{x e_1}{\sqrt{x^2 + y^2}} - y e_2 \\ J_f (e_2) &= \frac{y e_1}{\sqrt{x^2 + y^2}} + x e_2\end{align*}$$
Along the curve, $x = \cos t$ and $y = \sin t$, so we get
$$\begin{align*} J_f (\dot \ell(t)) &= -(\sin t )(e_1 \cos t - e_2 \sin t) + (\cos t)(e_1 \sin t + e_2 \cos t) \\ &= e_2\end{align*}$$
Remember that $e_2$ is associated with $\theta$--this says that, unsurprisingly, the velocity is entirely in the $\theta$ direction along this curve. We conclude that $\dot \ell(t) = J_f^{-1}(e_2) = e_\theta$.
In conclusion, we started with a tangent vector $\dot \ell(t)$ in our Cartesian coordinate system, and we moved it--using the Jacobian $J_f$--into a deformed plane where $(r,\theta)$ are "Cartesian" coordinates instead. The Jacobian is what moves tangent vectors from one space to another (or between coordinate systems), but positions are different and will always be handled by the full, nonlinear transformation.
One way you can remember this is that the Jacobian is like the derivative of the transformation, and so it's appropriate for moving things involving derivatives, like $\dot \ell(t)$, which is a velocity.
Good question. And you're right that we authors of elementary differential geometry texts don't always include these details.
Without loss of generality, you can assume the point in question is the origin, and then you have
$$\alpha(s)=-\lambda(s)T(s),$$
where $T(s)$ is the unit tangent vector. (We may always parametrize by arclength, but in this case it's not really relevant.)
Now $\alpha$ and $T$ are smooth functions, so therefore
$$\lambda = -\langle\alpha,T\rangle$$
is smooth as well. (If you want to be entirely pedantic, you can prove by the product rule and induction that if $f,g$ are $C^k$ functions, then $\langle f,g\rangle$ is also $C^k$. Or you can write it out in terms of component functions and then it's just a sum of products of $C^k$ functions.)
Similar arguments pertain when you write linear combinations of all the vectors in the Frenet frame. E.g., if $\alpha$ is a smooth curve with $\kappa\ne 0$ and you write
$$\alpha = \lambda T+\mu N + \nu B,$$
then the coefficient functions will similarly all be smooth.
Best Answer
$\newcommand{\R}{\mathbb{R}}$
Locally, any curve in $\R^2$ is a graph of some function $f \colon \R \to \R$, i.e., as the set $\Gamma = \{ (x,f(x)): x \in \R \}$. To make it straight, one only needs a very simple change of coordinates: $$ \Phi(x,y) = (x,y-f(x)), \quad \Phi^{-1}(x,y) = (x,y+f(x)). $$ It's simple to check that $\Phi$ transform $\Gamma$ into the line $(x,0)$.
Of course it only works locally, and the circle is a good reason why.