From the very scratch.
I can only confirm what Ninad Munshi was saying
that you are asking for problems when you do vector integrals with non Cartesian unit vectors.
To disentangle the notation I write
$$
\boldsymbol{\gamma}(t)=\gamma_x(t)\,\hat{\boldsymbol{x}}+\gamma_y(t)\,\hat{\boldsymbol{y}}
$$
for the curve over which we integrate and start with Cartesian coordinates. The tangent vector at this curve is
$$
\boldsymbol{\gamma}'(t)=\gamma_x'(t)\,\hat{\boldsymbol{x}}+\gamma_y'(t)\,\hat{\boldsymbol{y}}=\underbrace{\gamma_x'(t)\,\partial_x+\gamma_y'(t)\,\partial_y}\,.
$$
The notation in the underbraced term has the huge advantage that
the chain rule leads directly to the expressions in polar coordinates:
$$
\boldsymbol{\gamma}'(t)=\gamma_r'(t)\,\partial_r+\gamma_\theta'(t)\,\partial_\theta\,,
$$
where
\begin{align}\tag{1}
\partial_x&=\cos\theta\,\partial_r-\color{red}{\tfrac1r}\sin\theta\,\partial_\theta\,,&
\partial_y&=\sin\theta\,\partial_r+\color{red}{\tfrac1r}\cos\theta\,\partial_\theta\,,\\[2mm]
\gamma_x'&=\cos\theta\,\gamma_r'-\color{red}{r}\sin\theta\,\gamma_\theta'\,,&
\gamma_y'&=\sin\theta\,\gamma_r'+\color{red}{r}\cos\theta\,\gamma_\theta'\,.\tag{2}
\end{align}
The common physics convention is that the normalized unit basis vectors are
\begin{align}
\hat{\boldsymbol{r}}&=\partial_r\,,&
\hat{\boldsymbol{\theta}}&=\color{red}{\tfrac1r}\partial_\theta\,.
\end{align}
By this normalization (1) and (2) become orthonormal transformations.
The transformation (2) holds for any vector field.
When it is made clear that its polar components are to be understood
w.r.t. the coordinate basis $\{\partial_r,\partial_\theta\}=\{\hat{\boldsymbol{r}},\color{red}{r}\hat{\boldsymbol{\theta}}\}$ this becomes
\begin{align}\tag{3}
F_x&=\cos\theta\,F_r-\color{red}{r}\sin\theta\,F_\theta\,,&
F_y&=\sin\theta\,F_r+\color{red}{r}\cos\theta\,F_\theta\,.
\end{align}
The vector field that we integrate along $\gamma$ is
\begin{align}
\boldsymbol{F}=F_x\,\hat{\boldsymbol{x}}+F_y\,\hat{\boldsymbol{y}}=
F_x\,\partial_x+F_y\,\partial_y=
F_r\,\partial_r+F_\theta\,\partial_\theta=F_r\,\hat{\boldsymbol{r}}+\color{red}{r}F_\theta\,\hat{\boldsymbol{\theta}}\,,
\end{align}
where the polar notation is compatible with (1) and (2) and the fact that we express $F_r,F_\theta$ in the coordinate basis.
In Cartesian coordinates, the line integral is
\begin{align}
\int_C\boldsymbol{F}\cdot ds=\int_a^bF_x\gamma_x'+F_y\gamma_y'\,dt\,.%=\int_a^bF_r\,\gamma_r'+F_\theta\gamma_\theta'\,dt\,
\end{align}
where
$$
\boldsymbol{\gamma}(t)=\pmatrix{\gamma_x\\\gamma_y}=\pmatrix{r\cos t\\r\sin t}
$$
(these are Cartesian coordinates!). Therefore,
$$
\boldsymbol{\gamma}'(t)=\pmatrix{\gamma_x'\\\gamma_y'}=\pmatrix{-r\sin t\\r\cos t}\,.
$$
In your case $\boldsymbol{F}=r\hat{\boldsymbol{\theta}}=\partial_\theta$ so that $F_r=0$ and $F_\theta=\color{red}{1}\,.$ Using
(3) we get
$$
F_x=-\color{red}{r}\sin\theta\,,\quad F_y=\color{red}{r}\cos\theta\,.
$$
Choosing $t=\theta$ in $[0,2\pi]$ we get
\begin{align}
\int_C\boldsymbol{F}\cdot ds=\int_0^{2\pi}\color{red}{r^2}\sin^2\theta+\color{red}{r^2}\cos^2\theta=2\pi\color{red}{r^2}\,.
\end{align}
Perhaps it is worth mentioning that
$$
F_x=-r\sin\theta\,,\quad F_y=r\cos\theta\,,\quad \boldsymbol{F}=\pmatrix{-y\\x}
$$
so that the line integral around the circle coincides with that of the
one-form $-y\,dx+x\,dy\,.$ Since the radius is $r$ this form
coincides on $C$ with
$$
\boldsymbol{\omega}=r^2\frac{-y\,dx+x\,dy}{x^2+y^2}
$$
which is known to equal $r^2\,d\theta$ and has a very simple integral. No headaches whasoever.
