Riemannian Metric definition, Definition 2.1. chapter 1 Do Carmo Riemannian Geometry This is probably a stupid question. The book in the title defines

Definition 2.1. A Riemannian metric on a differentiable manifold $M$ is a correspondence which associates to each point $p$ of $M$ an inner product $\left\langle \cdot,\cdot \right\rangle_p$ (that is symmetric, bilinear, positive definite form) on the tangent space $T_p M$, which varies differentiably in the following sense: If $x : U \subset \mathbb{R}^n \to M$ is a system of coordinates around $p$, with $x(x_1,\ldots,x_n) = q \in x(U)$ and $\frac{\partial}{\partial x_i}(q) = dx_q (0,\ldots,1,\ldots,0)$, then $\left\langle \frac{\partial}{\partial x_i}(q),\frac{\partial}{\partial x_j}(q)\right\rangle_q = g_{ij}(x_1,\ldots,x_n)$ is a differentiable function in $U$

The first issue is the exact meaning of
$$
\frac{\partial}{\partial x_i} (q) = dx_q (0,\ldots,1,\ldots, 0)
$$
In general the differential of a map is the Jacobian of the coordinate transformation. I.e. if $\varphi$ is a differentiable map between two differentiable manifolds then it's differential is a linear map between the two tangent spaces of such manifolds. This map is characterized by the jacobian matrix of the coordinate transformation.
I would assume that $d x_q$ then is simply the velocity vector at $q$, does the notation $dx_q(0,\ldots,1,\ldots 0)$ means simply the $i-th$ row or column of this matrix?
The second issue...
It is my understanding that $\left\langle \frac{\partial}{\partial x_i}(q),\frac{\partial}{\partial x_i}(q)\right\rangle_q$ is an inner product on the tangent space $T_q M$. Two generic elements $X,Y \in T_q M$ generic element of such space would be given by
$$
\begin{array}{l}
X = \sum_{i=1}^n a_i(q) \frac{\partial}{\partial x_i} \\
Y = \sum_{i=1}^n b_i(q) \frac{\partial}{\partial x_i}
\end{array}
$$
However in a more simple setting (like regular surfaces) the Riemannian metric degenerates simply in the first fundamental form, however I do struggle to put the bits of the definition together to show this.
The reason of this question is that I'd like to assume I have some $X,Y \in T_p M$ how do I use the definition of Riemannian metric to compute the dot product?
 A: Fellow student of a Riemannian Geometry course here.
First off, Do Carmo is a poor textbook for a first introduction. I'd recommend reading another more accessible and thorough textbook first.
The differential of a map in differential geometry (of which Riemannian Geometry is a subgeometry) is a derived map on the tangent vectors. Intuitively, you can define it as follows. Let $\gamma$ be a curve in $M$ with velocity vector $\vec{v}$ at $m$, and let $f: M \to N$ be a smooth map, then the differential of $f$ is the smooth map which maps $\vec{v}$ to the velocity vector of $f \circ \gamma$ at $f(m)$.
If the map is a diffeomorphism (charts are), the differential will map a base of the tangent space $T_m M$ to a basis of the tangent space at $T_{f(m)} N$.
So $d \mathbf{x}_q$ is a map from the tangent space of $\mathbb{R}^n$ at $x^{-1}(q)$ (which is just $\mathbb{R}^n$ again), to the tangent space of $M$ at $q$. You're mapping one of the canonical base vectors $(0, \ldots, 1, \ldots, 0)$ to a corresponding tangent vector in your manifold. As you pointed out, there is a close relation to the Jacobian; the Jacobian is a matrix, and therefore it maps vectors to vectors.
Your interpretation as an inner product is correct. I do not understand what you mean by the first fundamental form, however. Do you mean to use the standard Euclidian product defined on $\mathbb{R}^3$ to define a Riemannian metric on a regular space in $\mathbb{R}^3$?
A: *

*Basically, do Carmo is denoting by  $x:U\to M$ a coordinate map. In my opinion, this  is somewhat confusing notation, so let's use $\phi:U\to M$ instead.  What he is saying is that if $q$ is the image of $p\in U$, i.e. $q=\phi(p)$, then by  definition $d\phi_p: T_p U=\mathbb{R}^n\to T_qM$. Then, if $e_i$ is the $i^{th}$ standard basis vector on $\mathbb{R}^n$ for $1\le i \le n$, the  definition of $\frac{\partial}{\partial x_i}(q)$ is exactly 
$$ \frac{\partial}{\partial x_i}\bigg|_q=d\phi_p(e_i).$$
So, associated to a coordinate system $(x_1,\ldots, x_n)$ on $M$, we get by taking the images of the  standard basis vectors in $\mathbb{R}^n$ a basis for $T_qM$, namely these $\partial_{x_i}(q)$ operators. 

*In the context of surfaces $M\subseteq \mathbb{R}^3$, the first fundamental form is typically defined as a bilinear form $I:T_qM\times T_qM\to T_qM$ so that with $v,w\in T_qM$, $I(v,w)=\langle v,w\rangle_M$ where $\langle v,w\rangle_M$ denotes the restriction of the dot product $\langle\:\cdot,\cdot\:\rangle_{\mathbb{R}^3}$ on $\mathbb{R}^3$ to $M$. Classically, the matrix representation of $I$ with respect to a basis $v_q,w_q$ of $T_qM$ is 
$$ \begin{bmatrix}
E(q)&F(q)\\
F(q)&G(q)
\end{bmatrix}$$
where (as for any symmetric bilinear form) $E(q)=\langle v_q,v_q\rangle$, $F(q)=\langle v_q,w_q\rangle=\langle w_q,v_q\rangle$, and $G(q)=\langle w_q,w_q\rangle$. So, if we were to use this bilinear form as the metric of $M\subseteq \mathbb{R}^3$ an embedded surface, taking $(x_1,x_2)$ as coordinates near $q\in M$, we would get associated partial derivatives $(\partial_1,\partial_2)_q$ and then in terms of these coordinates, $E(q)=\langle \partial_1(q),\partial_1(q)\rangle_q=g_{11}(q)$, 
$$F(q)=\langle \partial_1(q),\partial_2(q)\rangle_q=\partial_2(q),\partial_1(q)\rangle_q=g_{12}(q)=g_{21}(q)$$
and $G(q)=\langle \partial_2(q),\partial_2(q)\rangle_q=g_{22}(q)$. So, the matrix of the first fundamental form is 
$$ 
\begin{bmatrix}
g_{11}(q)&g_{12}(q)\\
g_{12}(q)&g_{22}(q)
\end{bmatrix}$$
in terms of our new notion of $g_{ij}$. Of course, the case of $M\subseteq \mathbb{R}^3$ is special, because  the choice of embedding $i:M\hookrightarrow{} \mathbb{R}^3$ endows the surface with  a metric inherited from $\mathbb{R}^3$.

*To compute inner products of tangent vectors on a Riemannian manifold (or of vector fields), fix local coordinates $(x_1,\ldots, x_n)$ around a point $q$ of interest. Then compute the $g_{ij}(q)$. Knowing this information completely determines the quadratic form of the inner product, which is then given by a matrix $(g_{ij})$
$$ 
\begin{bmatrix}
g_{11}&\cdots& g_{1n}\\
\vdots&\ddots&\vdots\\
g_{n1}&\cdots& g_{nn}
\end{bmatrix}. $$ 
Now, if in terms of the tangent vectors $(\partial_1(q),\ldots, \partial_n(q))$, and $v,w\in T_qM$, 
$$ w=
\begin{bmatrix}
w_1\\
\vdots\\
w_n
\end{bmatrix},\:\:\:\:\:
v=
\begin{bmatrix}
v_1\\
\vdots\\
v_n
\end{bmatrix}.$$
Then $\langle v,w\rangle_q=v^T(g_{ij})w$ as normal matrix multiplication.
