How does a metric tensor describe geometry on a manifold? I’m fairly new to the subject to the area of differential geometry, but as I understand it, the metric tensor $g$ is a tensor field that acts on the tangent space $T_{p}M$ to each point $p$ on a (Riemannian) manifold $M$. Specifically, it is defined as a mapping $g:T_{p}M\times T_{p}M\rightarrow\mathbb{R}$ such that $(v,w)\mapsto g(v,w)$ where $v,w \in T_{p}M$. The metric tensor intuitively gives the inner product of two vectors in a vector space, and thus can be used to determine magnitudes of vectors as well as the angle between intersecting curves tangent to two tangent vectors at a given point.
What I’m really unsure about is how the metric actually describes geometry on the manifold $M$? I know that one can choose a set of basis vectors adapted to a given set of coordinates on the manifold, a coordinate basis, such that the metric takes the form $$g=g_{\mu\nu}(x)dx^{\mu}\otimes dx^{\nu}\equiv g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$$ where $g_{\mu\nu}(x)=g\left(\frac{\partial}{\partial x^{\mu}}, \frac{\partial}{\partial x^{\nu}}\right)$. But this just gives the metric tensor at a point. How can one use $g_{\mu\nu}(x)$ in the entire coordinate chart? Or is the point that one evaluates $g_{\mu\nu}(x)$ at each point lying within the domain of the coordinate chart?
 A: 
What I’m really unsure about is how the metric actually describes geometry on the manifold $M$?

Geometry is nothing but measuring distances within the manifold, and the metric tells you how to do that. Once you have a ruler, you can do integrals, measure curvature and etc..

I know that one can choose a set of basis vectors adapted to a given set of coordinates on the manifold, a coordinate basis, such that the metric takes the form $$g=g_{\mu\nu}(x)dx^{\mu}\otimes dx^{\nu}\equiv g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$$ where $g_{\mu\nu}(x)=g\left(\frac{\partial}{\partial x^{\mu}}, \frac{\partial}{\partial x^{\nu}}\right)$. But this just gives the metric tensor at a point. How can one use $g_{\mu\nu}(x)$ in the entire coordinate chart? Or is the point that one evaluates $g_{\mu\nu}(x)$ at each point lying within the domain of the coordinate chart?

You can evaluate the metric tensor at each point in the domain of your coordinate chart. If you have an atlas of charts that covers the entire manifold, then you can evaluate the metric entirely.
EDIT: I'll give an example that I hope will solve your doubts in the comments. Lets look at the manifold $S^{2}$ embedded in $\mathbb{R}^{3}$. It can be described by the chart
$$\Phi:\left[0,\pi\right]\times\left[0,2\pi\right)\rightarrow S^{2}$$
$$\Phi\left(\theta,\varphi\right)=\begin{pmatrix}\sin\theta\cos\varphi\\ \sin\theta\sin\varphi\\ \cos\theta\end{pmatrix}$$
You can check the Jacobian of this chart and see that it is singular at the poles, but this can be neglected since those are just two points. Lets now calculate the metric. You have
$$\frac{\partial\Phi}{\partial\theta}=\begin{pmatrix}\cos\theta\cos\varphi\\ \cos\theta\sin\varphi\\ -\sin\theta\end{pmatrix}$$
$$\frac{\partial\Phi}{\partial\varphi}=\begin{pmatrix}-\sin\theta\sin\varphi\\ \sin\theta\cos\varphi\\ 0\end{pmatrix}$$
so the metric is given by
$$g_{\mu\nu}=\begin{pmatrix}\left<\frac{\partial\Phi}{\partial\theta},\frac{\partial\Phi}{\partial\theta}\right>&\left<\frac{\partial\Phi}{\partial\theta},\frac{\partial\Phi}{\partial\varphi}\right>\\\left<\frac{\partial\Phi}{\partial\varphi},\frac{\partial\Phi}{\partial\theta}\right>&\left<\frac{\partial\Phi}{\partial\varphi},\frac{\partial\Phi}{\partial\varphi}\right>\end{pmatrix}=\begin{pmatrix}1&0\\0&\sin^{2}\theta\end{pmatrix}$$
Does the explicit calculation help?
