differential geometry - Trouble finding the (inverse) transformations between covariant and contravariant components

Consider a point in cartesian and polar coordinates, $P = \{x,y\} = \{r,\theta\}$, such that $x = r \cos\theta$ and $y = r\sin\theta$. I have an arbitrary vector (e.g. a velocity vector) at point $P$, and I know it's cooridnates in cartesian coordinates, $\vec{v} = a^i e_i = a^0 e_0 + a^1 e_1$, where $e_i = \{\hat{x}, \hat{y}\}$, and I want to find the components in a bases that is the polar tangent vectors, $e'_i = \{\hat{r},\hat{\theta}\}$, i.e. $\vec{v} = a^{\prime i} e'_i = a^{\prime 0} e'_0 + a^{\prime 1} e_1$.

(Ultimately I need to do this exercise in 3D spherical coordinates, and do a couple of coordinate rotations in between, but I think this case encompasses my [current] confusion).

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Attempt:

I can define a transformation between coordinates based on, $$e'_j = e_i \, {S^i}_j = e_i \frac{\partial e'_j}{\partial e_i},$$

for example, $e'_0 = \hat{r} = \frac{\partial}{\partial r} = \frac{\partial}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial}{\partial y} \frac{\partial y}{\partial r}.$

Thus the tensor that transforms the bases $\{\hat{x},\hat{y}\} \rightarrow \{\hat{r},\hat{\theta}\}$, can be calculated as

$${S^i}_j = \pmatrix{\frac{\partial e'_0}{\partial e_0} & \frac{\partial e'_1}{\partial e_0} \\ \frac{\partial e'_0}{\partial e_1} & \frac{\partial e'_1}{\partial e_1}} = \pmatrix{\frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y}} = \pmatrix{\cos\theta & -\frac{1}{r} \sin\theta \\ \sin\theta & \frac{1}{r} \cos\theta},$$

and the values of $r,\theta$ here are, again, fixed based on the point $P$. Using this, we immediately get, $e'_i = \{\hat{r}, \hat{\theta} \} = \{\hat{x}\cos\theta + \hat{y}\sin\theta, \,\, - (\hat{x}/r)\sin\theta + (\hat{y}/r) \cos\theta \}$, which matches what I get from geometry. So this seems good.

The components of the vectors transform contravariantly, and so the transformation should be given by, ${T^i}_j = \partial e_j \, / \, \partial e'_i$. Going through the same procedure, I find the components of the tensor that transforms the coordinates $\{a_0,a_1\} \rightarrow \{a'_0, a'_1\}$ as,

$${T^i}_j = \pmatrix{\frac{\partial e_0}{\partial e'_0} & \frac{\partial e_1}{\partial e'_0} \\ \frac{\partial e_0}{\partial e'_1} & \frac{\partial e_1}{\partial e'_1}} = \pmatrix{\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}} = \pmatrix{\cos\theta & \sin\theta \\ -r \sin\theta & r \cos\theta}.$$

Now if we try this, we immediately get a problem: $a^{\prime i} = \{x\cos\theta + y\sin\theta, \,\, -rx\sin\theta + ry\cos\theta \}$. The 0th component seems fine, but the 1st has the wrong dimensions: length squared instead of dimensionless!

I also notice that while $T S = \mathbb{I}$ (the identity matrix), and thus $T = S^{-1}$, also $S T \neq \mathbb{I}$, i.e. $S \neq T^{-1}$...

Am I missing a transpose somewhere? Or am I assuming an orthonormality somewhere where it doesn't exist? Any help or pointers (even on improper terminology / syntax) is much appreciated!

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Edit: Rows-vs-columns, and left-vs-right matrix multiplication has always confused me, so I'm primarily following the indices and their positions, but I've still tried to be consistent with rows and columns. One "reason" to add a transpose, however, could be that while the inverse matrix should transform the components (as apposed to bases), the components are 'columns' while the bases 'rows', so perhaps that adds a transpose? But where does this come from in index notation?