Inner product on tangent space and metric tensor In our class we talked about integrating on submanifolds and as a short side remark our teacher told us that by knowing the metric tensor, it is possible to define an inner product on a tangent space and then he said: $\alpha(t):=\phi(tb)$, where $t$ is a real number, phi is a chart and b a vector in a lower-dimensional $\mathbb{R}^n$.
And then we said that $||\dot{\alpha}(0)||_2^2=b^T (g_{ij})b$.
My problem is the following: I see that the tangent space contains all those curves, but I do not see the idea behind this vector $b$. Especially, why is it meaningful to say $\alpha(t):=\phi(tb)$.
Could anybody explain to me the idea behind this definition? Or maybe a good reference that uses this approach, may also be helpful. Unfortunately, I could not find something similar so far.
I'm somewhat sceptical about the circumstance in the given answer, how one can define b.
 A: If I am understanding this correctly, I believe he is trying to define an inner product on a tangent space to construct a metric that comes explicitly from the chart structure on the manifold. 
One way to define the tangent space is as the space of vectors at your point which are the derivatives of curves going through your point. Specifically, suppose your manifold is $M$ and $\phi:U \to M$ is a chart with $U \subseteq \mathbb{R}^n$. Then the tangent space at $\phi(0)$ is exactly the set of vectors given by $(\phi \circ \alpha)'(0)$ where $\alpha$ is some curve in $U$ with $\alpha(0) = 0$. With this in mind we can understand the situation above.
He is saying for any vector $v$ in the tangent space $T_{\phi(0)}$ at $\phi(0)$, from the characterization above, there exists some curve (actually just a linear subspace) in $\mathbb{R}^n$ defined $\beta(t) = tb$ where $b$ is a vector in $U$, such that $v$ is given by $\alpha'(0)$ where $\alpha = \phi\circ \beta$ is the curve on $M$ defined by $\beta$.  Then we can define the norm of $v$ by
$$
\langle v,v \rangle = ||\alpha'(0)||^2_2 := b^T(g_{ij})b
$$
By linearity, this lets us define $\langle -,-\rangle$ for any two vectors $a$ and $b$ in $T_{\phi(0)}$ by using the above formula applied to $a+b$. Thus we have constructed an inner product on $T_{\phi(0)}$ directly from the chart $\phi$. The metric tensor $g_{ij}$ is compatible with charts and varies continually so we get a well defined (independent of charts) and continuously varying inner product on the tangent spaces. 
Edit: Here is one way to see that $b$ is uniquely defined. We get a basis of $T_{\phi(0)}$ given by the standard basis $e_1, \ldots, e_n$ of $\mathbb{R}^n$ and our parametrization $\phi$. Specifically, the vectors 
$$
b_i := d\varphi_0(e_i) = \frac{d}{dt}(\varphi(e_i t))\big|_0.
$$
This is a basis of the tangent space because the tangent space by definition is the $d\varphi_0(\mathbb{R}^n)$, i.e., the pushforward of the tangent space of our parametrization by the differential of the parametrization. The RHS of this equation is one way of calculating the differential $d\varphi_0(e_i)$, the other being just calculating the Jacobian matrix of $\varphi$ and applying it to $e_i$. Now, since the $b_i$ form a basis of $T_{\varphi(0)}$, we have that for any $v \in T_{\varphi(0)}$, $v$ can be expressed uniquely as
$$
v = \sum a_i b_i
$$
Since $b_i$ form a basis. Thus, take $b$ to be the vector 
$$
b = \sum a_i e_i.
$$
This maps to $v$ and is unique by construction since basis expressions are unique. Another way to see uniqueness is that $\varphi$ is a parametrization so $d\varphi_0$ is injective so if another vector say $b'$ mapped to $v$, we'd have $d\varphi_0(b) = d\varphi_0(b')$ so $b = b'$ by injectivity. 
