Trying to understand polar coordinate vectors

I'm trying to understand what the unit polar coordinate vectors (I'll denote them $\hat r$ and $\hat \phi$) are and if they form a basis for $\Bbb R^2$.

So from what I understand, $\hat r$ points radially from the center and $\hat {\theta}$ points orthogonally to $\hat r$, as in the picture below:

Then a vector $\vec v$ could be represented by a linear combination of $\vec v$, i.e. $\vec v=a \hat r + b \hat \phi$. My question is, if $\{\hat r,\hat \phi\}$ is a basis of $\Bbb R^2$, then knowing $(a,b)$ should completely specify any point, right? But I don't understand how either $\hat r$ or $\hat \phi$ specify the angle that the vector makes with the positive $x$-axis (or whatever that reference line is called in polar coordinates). I can see how, if you know the Cartesian representation of $\hat r$ or $\hat \phi$, you can then determine the other and thus specify any point, but can you specify any point in the plane without a conversion to Cartesian first? If not, why are $\hat r$ and $\hat \phi$ even useful?

Edit:
Muphrid does a good job of explaining why trying to think of $\hat r$ and $\hat \phi$ as basis vectors of $\Bbb R^2$ isn't the best way to use them. But in case anyone who reads this is interested in where they come from, how to use them, and how to derive the relations for the grad, div, and curl, I found this pretty good set of lecture notes.

Well, your vectors $\hat r$ and $\hat\phi$, as the picture greatly shows, form a basis.

The polar coordinates of a point are just $r$ and $\phi$: the distance from origin, and the angle enclosed with the positive wing of the $x$-axis.

If you consider the length of your vectors as $|\hat r|=|\hat\phi|=r\ $ then these vectors are just the $x$ and $y$ directional derivatives of the function $\Bbb R^2\to\Bbb R^2$ $$(r,\phi)\ \mapsto\ (r\cos\phi,\,r\sin\phi)$$ i.e. which takes $(r,\phi)$ to the point it represents as polar coordinates (the point in your picture).

The basis vectors $\hat r, \hat \phi$ are vector fields. It happens to be true that, for any point described by $(r_0, \phi_0)$, then a vector describing that position is equal to $r_0\hat r(r_0, \phi_0)$. That...may only be marginally useful.

Nevertheless, given a point $(r_0, \phi_0)$, the values of those vector fields at that point--i.e. $\hat r(r_0, \phi_0)$ and $\hat \phi(r_0, \phi_0)$--span a 2d vector space. This is called the tangent space at the point $(r_0, \phi_0)$. This is not suitable for describing positions (see note), but it is suitable for describing, among other things, the vectors tangent to curves that go through the point $(r_0, \phi_0)$.

Note: you could, of course, choose some arbitrary $(r_0, \phi_0)$ to evaluate $\hat r, \hat \phi$, and then those values span $\mathbb R^2$, but the choice of a point there is arbitrary and not related to any particular position you want to describe.

On the topic of Newton's 2$^{\text{nd}}$ from Cartesian to polar. Converting from my $\LaTeX$ notes to mathjax is a pain.

In Cartesian coordinates, we can write the vector $\mathbf{r}$ as $x\hat{\mathbf{x}} + y\hat{\mathbf{y}}$ where $x = r\cos(\phi)$, $y = r\sin(\phi)$, $\lVert \mathbf{r}\rVert = \sqrt{x^2 + y^2}$, and $\phi = \arctan\left(\frac{y}{x}\right)$ see figure.

Since we have two independent unit vectors, we can write our force as a linear combination of the basis vectors $\{\hat{\mathbf{r}}, \pmb{\phi}\}$. $$ \mathbf{F} = F_r\hat{\mathbf{r}} + F_{\phi}\pmb{\phi} \tag{1} $$

What is $\dot{\mathbf{r}}$?

\begin{align} \dot{\mathbf{r}} &= \frac{d}{dt}\left( \lVert\mathbf{r}\rVert\hat{\mathbf{r}}\right)\\ &= \dot{r}\hat{\mathbf{r}} + r\frac{d\hat{\mathbf{r}}}{dt}\tag{2} \end{align}

Let $\Delta t = t_2 - t_1 \ll 1$. From the figure directly above, we have that $\Delta\mathbf{r} = \Delta\phi\pmb{\phi}$. \begin{align*} \Delta\mathbf{r} &= \Delta\phi\pmb{\phi}\frac{\Delta t}{\Delta t}\\ &= \dot{\phi}\Delta t\pmb{\phi} \tag{since \(\frac{\Delta\phi}{\Delta t} = \dot{\phi}\)}\\ \frac{\Delta\mathbf{r}}{\Delta t} &= \dot{\phi}\pmb{\phi}\tag{3} \end{align*} Now we can take the limit of equation (3) as $\Delta t\to 0$. Thus, we have that \begin{align} \lim\limits_{\Delta t\to 0}\frac{\Delta\mathbf{r}}{\Delta t} &= \dot{\phi}\pmb{\phi}\\ \frac{d\hat{\mathbf{r}}}{dt} &= \dot{\phi}\pmb{\phi}\tag{4} \end{align} Now we can plug equation (4) into equation (2) to obtain \begin{align} \dot{\mathbf{r}} &= \dot{r}\hat{\mathbf{r}} + r\dot{\phi}\pmb{\phi}. \tag{5} \end{align} Since $\dot{\mathbf{r}} = \mathbf{v}$, we can write equation (5) in component form of velocity, $\mathbf{v} = v_r\hat{\mathbf{r}} + v_{\phi}\pmb{\phi}$. Then $v_r = \dot{r}$ and $v_{\phi} = r\dot{\phi}$. Note that $\omega$ is used instead of $\dot{\phi}$ in many textbooks as well. We will also be using $\omega$ and $\dot{\phi}$ interchangeable throughout the notes. In order to define Newton's $2^{\text{2nd}}$ Law in polar coordinates, we still need to determine $\mathbf{a} = \ddot{\mathbf{r}}$. \begin{align*} \mathbf{a} &= \ddot{\mathbf{r}}\\ &= \frac{d}{dt}\dot{\mathbf{r}}\\ &= \frac{d}{dt}\left(\dot{r}\hat{\mathbf{r}} + r\dot{\phi}\pmb{\phi}\right)\\ &= \ddot{r}\hat{\mathbf{r}} + \dot{r}\frac{d\hat{\mathbf{r}}}{dt} + \dot{r}\dot{\phi}\pmb{\phi} + r\ddot{\phi}\pmb{\phi} + r\dot{\phi}\frac{d\pmb{\phi}}{dt}\\ &= \ddot{r}\hat{\mathbf{r}} + 2\dot{r}\dot{\phi}\pmb{\phi} + r\ddot{\phi}\pmb{\phi} + r\dot{\phi}\frac{d\pmb{\phi}}{dt}\tag{6} \end{align*}

From figure above, we see that $\Delta\pmb{\phi} \approx \Delta\phi\hat{\mathbf{r}}$. \begin{align*} \Delta\pmb{\phi} &\approx \Delta\phi\hat{\mathbf{r}}\\ &= \Delta\phi\hat{\mathbf{r}}\frac{\Delta t}{\Delta t}\\ &= -\dot{\phi}\hat{\mathbf{r}}\Delta t \tag{since $\frac{\Delta\phi}{\Delta t} = \dot{\phi}$}\\ \frac{\Delta\pmb{\phi}}{\Delta t} &= -\dot{\phi}\hat{\mathbf{r}}\tag{7} \end{align*} Taking the limit of equation (7) as $\Delta t\to 0$, we have \begin{align} \lim\limits_{\Delta t\to 0} \frac{\Delta\pmb{\phi}}{\Delta t} &= -\dot{\phi}\hat{\mathbf{r}}\\ \frac{d\pmb{\phi}}{dt} &= -\dot{\phi}\hat{\mathbf{r}}\tag{8} \end{align} Plugging in equation (8) into equation (6) and grouping, we obtain $$ \ddot{\mathbf{r}} = \left(\ddot{r} - r\dot{\phi}^2\right)\hat{\mathbf{r}} + \left(2\dot{r}\dot{\phi} + r\ddot{\phi}\right)\pmb{\phi}. \tag{9} $$ Now we can write $\mathbf{F} = m\ddot{\mathbf{r}}$ in component form \begin{align} F_r &= m\left(\ddot{r} - r\dot{\phi}^2\right)\tag{10}\\ F_{\phi} &= \left(2\dot{r}\dot{\phi} + r\ddot{\phi}\right)\tag{11} \end{align} If the radius is constant (a circle), then $\dot{r} = \ddot{r} = 0$. For a circle, \begin{align} \frac{F_r}{m} &= -r\dot{\phi}^2\tag{angular velocity}\\ \frac{F_{\phi}}{m} &= r\ddot{\phi}\tag{angular acceleration} \end{align}