$\hat{\theta}$ tells you the angle you need to be at from the positive $x$-axis, and $\hat{r}$ tells you how far you need to walk out from the origin. The magnitude is most certainly given by the $\hat{r}$ component. Perhaps your issue is that you're not adding/multiplying vectors in polar form properly. You cannot simply take, for example, $v_1 = 1\hat{r} + \pi\hat{\theta}$, and conclude that $v_1 + v_1 = 2\hat{r} + 2\pi\hat{\theta}$. Notice that the angle has been changed, which shouldn't happen for two vectors that are colinear.
It is better to represent vectors as $re^{i\theta}$. From which in our example we have, $v_1 + v_1 = e^{i\pi} + e^{i\pi} = 2e^{i\pi}$, which has an $\hat{r}$ of 2, and $\hat{\theta}$ of still $\pi$.
It will turn out that the rotation in polar coordinates is nothing else than a shift
of coordinate functions by an angle $\alpha$. Note that we usually shift a function
$g$ on $\mathbb R$ by $\alpha$ by going to the function $x\mapsto g(x\color{red}{-}\alpha)$.
A very convenient method to formalize vector fields in different coordinates is to treat them as derivations. This is very common in differential geometry. In $2d$ Cartesian coordinates a vector field is then
$$
X=X_1\frac{\partial}{\partial x_1}+X_2\frac{\partial}{\partial x_2}
$$
where $X_1,X_2$ are scalar functions on $\mathbb R^2$.
From the Cartesian-polar transformation
\begin{align}
r&=\sqrt{x_1^2+x_2^2}\,,&
\theta&=\arctan\frac{x_2}{x_1}\,\\
x_1&=r\cos\theta\,,&x_2&=r\sin\theta
\end{align}
we get
\begin{align}
\frac{\partial x_1}{\partial r}&=\cos\theta\,,&\frac{\partial x_1}{\partial \theta}=-r\sin\theta\,,\\
\frac{\partial x_2}{\partial r}&=\sin\theta\,,&\frac{\partial x_2}{\partial \theta}=r\cos\theta\,.
\end{align}
By the chain rule we have for any differentiable function $f$ on $\mathbb R^2$
\begin{align}
\frac{\partial f}{\partial r}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial r}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial r}
=\frac{\partial f}{\partial x_1}\cos\theta+\frac{\partial f}{\partial x_2}\sin\theta\,,\\[3mm]
\frac{\partial f}{\partial \theta}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial \theta}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial \theta}
=-\frac{\partial f}{\partial x_1}r\sin\theta+r\frac{\partial f}{\partial x_2}\cos\theta\,.
\end{align}
In matrix notation and dropping the test function this is the transformation rule for the basis vector fields:
$$
\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}=\begin{pmatrix}\cos\theta &\sin\theta\\-r\sin\theta&r\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}.
$$
Clearly,
$$
\begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}=\begin{pmatrix}\cos\theta &-\frac{1}{r}\sin\theta\\\sin\theta&\frac{1}{r}\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}.
$$
The vector field $X$ in polar coordinates is therefore,
\begin{align}
X'&=X_1'\Big(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\Big)+X_2'\Big(\sin\theta\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\frac{\partial}{\partial \theta}\Big)\\[3mm]
&=\Big(X_1'\cos\theta+X_2'\sin\theta\Big)\frac{\partial}{\partial r}+
\Big(-X_1'\frac{1}{r}\sin\theta+X_2'\frac{1}{r}\cos\theta\Big)\frac{\partial}{\partial \theta}
\end{align}
where $X'_1(r,\theta)=X_1(x_2,x_2)$ and $X'_2(r,\theta)=X_2(x_2,x_2)\,.$
The whole point is now that nothing is simpler than rotating the vector field $X'$ by an angle $\alpha$ around the origin.
This rotation is a shift which gives
\begin{align}
X''
&=\Big(X_1''\cos(\theta-\alpha)+X_2'\sin(\theta-\alpha)\Big)\frac{\partial}{\partial r}+
\Big(-X_1''\frac{1}{r}\sin(\theta-\alpha)+X_2''\frac{1}{r}\cos(\theta-\alpha)\Big)\frac{\partial}{\partial \theta}
\end{align}
where $X''_1(r,\theta)=X'_1(r,\theta-\alpha)$ and $X''_2(r,\theta)=X_2'(r,\theta-\alpha)\,.$
Best Answer
You could do integration by parts
$$\int_0^{\frac{\pi}{2}} \theta \hat{\theta}d\theta = \theta \hat{r}\Bigr |_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \hat{r}d\theta = \frac{\pi}{2}\hat{r}\left(\frac{\pi}{2}\right) + \hat{\theta}\left(\frac{\pi}{2}\right) - \hat{\theta}(0)$$
But as you can see, we have stumbled on a fundamental problem of using polar unit vectors: they don't uniquely describe constant vectors and are entirely useless for doing so.
$$\hat{r}\left(\frac{\pi}{2}\right) = \hat{\theta}(0) = \hat{y}$$
Once you plug in numbers for the polar unit vectors, it's game over. There is no point in continuing to use polar coordinates. Polar unit vectors are better used for vector fields, whose values change at different points in space.