Riemannian metric on $\Bbb R^2$ in polar coordinates 
The Riemannian metric of the Euclidean space $\Bbb R^n$ is $g = \delta_{ij}dx^i dx^j$.
Write $g$ in the case of the plane $\Bbb R^2$ by using the polar coordinates $(r, \theta)$, $x^1 = r\cos \theta$, $x^2 = r \sin\theta$.

I'm learning about the Riemannian metric and I have some problems with the definitions. According to wikipedia the metric $g$ is defined as $$g_p: T_pM \times T_pM \to \Bbb R$$ which means that it's always defined with respect to some point $p$ on the manifold. In the question however it would look like it's defined without any "base point", what might be the reason for this?
Regarding the space $\Bbb R^2$ with the polar coordinates $(r, \theta)$ for each $p \in \Bbb R^2$ we have a basis for $T_p \Bbb R^2$ given by $\left\{\dfrac{\partial}{\partial r} \bigg|_p, \dfrac{\partial}{\partial \theta} \bigg|_p\right\}$ so can this be used to express the metric in polar coordinates?
 A: $dr$ and $d\theta$ are one forms, and $dr^2 $ is a common shortcut for $dr\otimes dr$ (similarly for $d\theta^2$). These are dual to the corresponding tangent vectors, in the sense that $dx^i(\frac{\partial}{dx^j})=\delta_i^j$.
The components of the metric in a given coordinate system $(x^1,\dots, x^n)$ are $g_{ij}= g(\frac{\partial}{dx^i},\frac{\partial}{dx^j})$. An equivalent way to write this down with forms is $g = g_{ij}dx^i\otimes dx^j$ - the $\otimes$ symbol is often omitted. Here, the Einstein summation convention is used.
Now for polar coordinates you usually write $r,\theta$ for $x^1, x^2$ If you want to reserve the $x^i$ for cartesian coordinates make that $\xi^1= r, \xi^2 = \theta$ or even $\xi^r= r, \xi^\theta = \theta$ (which I don't like because it is so strictly two dimensional).
In any case you get, with a notation for the indices which should now be self-explanatory, $$g = g_{rr}dr\otimes dr + g_{\theta \theta}d\theta \otimes d\theta$$ (the mixed terms turn out to vanish, which, of course, requires proof by calculation left to you, but see below). For calculations you want to find $g_{rr} $ and $g_{\theta \theta}$.
Now, e.g., $$g_{rr}= g(\frac{\partial }{\partial r}, \frac{\partial }{\partial r})
=g\left(\frac{\partial x^i}{\partial r}\frac{\partial}{\partial x^i},
\frac{\partial x^j}{\partial r}\frac{\partial}{\partial x^j}\right)=1$$
and
$$g_{\theta\theta}= g(\frac{\partial }{\partial \theta}, \frac{\partial }{\partial \theta})
=g\left(\frac{\partial x^i}{\partial \theta}\frac{\partial}{\partial x^i},
\frac{\partial x^j}{\partial \theta}\frac{\partial}{\partial x^j}\right)=r^2$$
(using the definitions $x^1 = r\cos(\theta), x^2 = r\sin(\theta)$, the bilinearity of $g$ and the fact that $g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})=\delta_{ij}$ and a short calculation), so $g$ becomes
$$ g = dr\otimes dr + r^2d\theta \otimes d\theta = dr^2 + r^2d\theta^2 $$ with the last term just abbreviating the one with the $\otimes$ notation.
(Sorry if this got a bit lengthy, I had to dig this out of some old memory. Hope it is correct ;-)
A: The metric maps two vectors to a number (distance), but $g_p$ is itself defined at all points $p$ here, thus $g_p$ is a function of a point $p(r_p, \theta_p)$.
