Operators in QM don't depend on coordinates, so changing coordinates does not affect them in the least; which is why your (2) is off. Recall how you handle Schroedinger's equation in polar coordinates. Stick to one dimension for simplicity.
The proper, operator, TDSE is
$$
\left (-i\hbar \partial_t + \frac{\hat p ^2}{2m} +V(\hat x) \right ) |\Psi(t)\rangle =0 \tag{1},
$$
which, dotted by $\langle x|$, yields your first TDSE in coordinate space,
$$
i \hbar \partial_t\langle x|\Psi(t)\rangle= \langle x| \frac{\hat p ^2}{2m} |\Psi(t)\rangle + \langle x|V(\hat x)|\Psi(t)\rangle \tag{1a}\\ i \hbar \partial_t \Psi( x, t)=-\tfrac{\hbar^{2}}{2 m} \partial_x ^{2} \Psi( x, t)+V(x) \Psi( x, t).$$
Under a coordinate transformation $ x\rightarrow x'= x'(x)$, this PDE is equivalent to
$$i \hbar \partial_t \Psi( x', t)=-\tfrac{\hbar^{2}}{2 m} \partial_{x'} ^{2} \Psi( x', t)+V(x') \Psi( x', t) ~.
\tag 2$$
The first term (kinetic) on the r.h.side can also be rewritten as
$$
-\tfrac{\hbar^{2}}{2 m} \partial_{x'} ^{2} \Psi( x', t) = -\tfrac{\hbar^{2}}{2 m} \left (\frac{1}{\partial_x x'(x)} ~ \partial_{x} \right )^{2} \Psi( x'(x), t) , \tag{3}
$$
on which you might practice with a simple example.
But this is exactly what you would have gotten by dotting (1) by $\langle x'(x)|$.
Edit as per comments, which have taken a left turn and point to a very different question, on canonical transformations. For the 2D polar coordinate question asked, however, the change of variable $(x,y) \to (r=\sqrt{x^2+y^2}, \phi =\arctan(y,x))$, (3) goes to the generalization
$$
-\tfrac{\hbar^{2}}{2 m}(\partial_{x} ^{2}+ \partial_{y} ^{2})= -\tfrac{\hbar^{2}}{2 m} \left (\partial_r^2+\frac{1}{r}\partial_r + \frac{1}{r^2}\partial_\phi^2 \right )^{2} ,
$$
and you have a plain change of coordinates.
As stated above,
$$
\hat x |r,\phi\rangle = r\cos\phi ~|r,\phi\rangle, ... \hbox{etc.}
$$
To be sure, you could define formal operators such as
$$
\hat r\equiv \sqrt{\hat x ^2 + \hat y ^2}, \implies ~~
\hat r |r,\phi\rangle= r|r,\phi\rangle,
$$
but, as indicated in the linked answer, operators of angular variables (skipped here) are beset with formal problems revolving around hermiticity, so some wise bypass them.
