Covariant and Contravariant Transformation - Example of Polar Coordinates

As @TedShifrin pointed out, the correct transformation for the dual is $(A^{-1})^T$.

For instance, in special relativity contravariant vectors transform as

$ V^{\mu '} = \Lambda ^{\mu '} _{\,\nu} V^{\nu}$,

where $ \Lambda ^{\mu '} _{\,\nu} $ is the Lorentz transformation taking component from the unprimed frame to the primed frame. For covariant components we have to use the inverse of the Lorentz transformation:

$ V_{\mu '} = \big( \Lambda^{-1} \big)^{\nu } _{\,\mu '} V_{\nu}$.

If we assume that $V'$ and $V$ are column vectors representing, $ V_{\mu '}$ and $V_{\nu}$, respectively. Then in matrix format we have:

$ V' = \big( \Lambda^{-1} \big)^T V$.

This can also be found here: https://en.wikipedia.org/wiki/Lorentz_transformation

Of course, one can get to the same conclusion in the context of group theory. If the vector $x$ transforms according to $x \rightarrow x'=Ax$ where $A$ is the group member. Then the dual $\tilde{x}$ transforms as $\tilde{x} \rightarrow \tilde{x} '= \big(A^{-1} \big)^T\tilde{x}$ so that $\tilde{x}^T x$ is an invariant in all frames.

I think that your answer is unnecessarily complicated for this question. In matrix notation equation (1) in the question is

\begin{equation} \begin{pmatrix} \hat{r} && \hat{\theta} \end{pmatrix} =\begin{pmatrix} \hat{x} && \hat{y} \end{pmatrix} \begin{pmatrix} cos\phi && -sin\phi \\ sin\phi && cos\phi \end{pmatrix} \triangleq \begin{pmatrix} \hat{x} && \hat{y} \end{pmatrix} M \end{equation}

and equation (3) expresses the transformation rule for a row (covariant) vector

\begin{equation} \begin{pmatrix} v_r && v_{\theta} \end{pmatrix}= \begin{pmatrix} v_x && v_y \end{pmatrix} M \end{equation}

so the transformation for a column (thus contravariant) vector is

\begin{equation} \begin{pmatrix} v_r \\ v_{\theta} \end{pmatrix} =M^T \begin{pmatrix}v_x\\ v_y\end{pmatrix} \end{equation}

Which is indeed the inverse of $M$ as the latter is orthogonal.