This is not a proper answer, but it might be useful anyway. It's just what I've found so far reading books and trying to make sense of everything I learned about vectors in both calculus and algebra.
You can geometrically picture vectors in $\mathbb{R}^n$ as arrows placed at the origin. Every vector can be uniquely expressed as a linear combination of $n$ linearly independent vectors, so for every basis of this vector space $(\vec{e}_1,\dots,\vec{e}_n)$ we can write:
$$
v=v^1 \vec{e}_1 + \cdots + v^n \vec{e}_n
$$
Although using a vector basis allows us to uniquely identify every vector in $\mathbb{R}^n$, it isn't really useful when trying to identify every "point" in $\mathbb{R}^n$ because of its limitations (your axes will have to be straight lines and will have to include the "canonical" origin).
In order to identify every point of $\mathbb{R}^n$ with more freedom, our first approach could be affine geometry. In $\mathbb{R}^n$ viewed as an affine space, we can define a coordinate system $(O,\mathcal{B})$, with $O$ a point in $\mathbb{R}^n$ and $\mathcal{B}$ a basis of $\mathbb{R}^n$ as a vector space. Axes are still straight lines, but now we can move from the origin to any other point $O$. This affine coordinate system is in a way a coordinate system, and it is definitely not the same as a basis — because $(O,\mathcal{B}) \neq \mathcal{B}$ — but we can do better.
We can define a system of $n$ equations (not necessarily linear, a system of linear equations would bring us back to the affine case) that uniquely identify every point in $\mathbb{R}^n$:
$$
x^i =\Phi^i(q^1,\dots,q^n), \space\space\space i=1,\dots,n
$$
This is precisely what we do when we define cylindrical or spherical coordinates: express $x,y,z$ in terms of three new variables ($\rho,\varphi,z$ and $r,\theta,\varphi$ respectively).
This means that the values $(q^1,\dots,q^n)$ will be our new coordinates. Note that, if this system were linear, we would need to require the coefficient matrix to have a non-zero determinant in order for this system to have a (unique) solution. For a general system of equations, by virtue of the implicit function theorem, the analogous condition is:
$$
\frac{\partial(x^1,\dots,x^n)}{\partial(q^1,\dots,q^n)} \neq 0
$$
i.e. the Jacobian of the transformation must be non-zero.
This new coordinates $(q^1,\dots,q^n)$ aren't related to any basis, but they induce one for every point $(q^1,\dots,q^n)$ in $\mathbb{R}^n$: the so-called coordinate basis of this coordinate system:
$$
\vec{v}_\mu = \frac{\partial\vec{\Phi}}{\partial q^\mu}
$$
In this sense, we can see than the components of the position vector at the point $p\in\mathbb{R}^n$ will be different from the coordinates of the point $p$ itself. The components depend on the coordinate basis (or any other basis which you define in terms of that one), while the coordinates of a point depend on the coordinate system itself.
In fact, now we're not talking about $\mathbb{R}^n$ as a vector space anymore, but this space does have a vector space attatched at every point, a vector space we call the tangent space at $p$: $T_p\mathbb{R}^n$. The coordinate basis at every point is the vector basis that our coordinate system induces for the tangent space at that point.
Note that the components of a vector (or a tensor, for that matter) may be called coordinates. I only see this when reading about pure vector spaces, without any sense of geometry, metrics or anything. For example, a matrix $A\in\mathcal{M}_{n\times n}$ which satisfies $\vec{v}_{\mathcal{B}'} = A\vec{v}_{\mathcal{B}}$ might be called a change of coordinates matrix (from the coordinates from $\mathcal{B}$ to the coordinates from $\mathcal{B}'$) or a change of basis matrix (from the basis $\mathcal{B}'$ to the basis $\mathcal{B}$, because it satisfies $\mathcal{B}'A=\mathcal{B}$, if we allow ourselves this abuse of notation).
Nonetheless, I always say components when referring to a vector (or tensor), and coordinates when referring to a point.
$\hat{\theta}$ 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 $\pi$.
Best Answer
The Polar Unit Vectors $e_{\theta}$ and $e_r$ are unit vectors like the Cartesian Unit Vectors $x$ and $y$. The difference is that $e_{\theta}$ and $e_r$ change direction with position.
To see how this works, consider the point (1,0): At (1,0) $e_r = x$ and $e_{\theta} = y$.
At the point (0,1): $e_r = y$ and $e_{\theta} = -x$.
You can think of the vectors by imaging a unit circle of radius 1. $e_r$ always points to a position on the circle while $e_{\theta}$ is tangent to the circle that points in in a counter-clockwise (increasing angle) direction.