Doubts about the Zorich explanation of coordinate systems and orientation classes I make a minimum of preamble by reporting what is written in Zorich, Mathematical Analysis II, 1st ed., Page 170-172:

Let $G$ and $D$ be diffeomorphic domains lying in two copies of the space
$\mathbb{R}^n$ endowed with Cartesian coordinates $(x^1,..., x^n)$ and $(t^1,..., t^n)$ respectively. A diffeomorphism $\varphi : D → G$ can be regarded as
the introduction of curvilinear coordinates $(t^1,..., t^n)$ into the
domain G via the rule $x = \varphi(t)$, that is, the point $x \in G$ is endowed
with the Cartesian coordinates $(t^1,..., t^n)$ of the point $t = \varphi^{-1}(x) \in D$. If we consider a frame $\mathbf{e}_1,...,\mathbf{e}_n$ of the tangent space $T\mathbb{R}^n_t$ at each
point $t \in D$ composed of the unit vectors along the coordinate
directions, a field of frames arises in $D$, which can be regarded as
the translations of the orthogonal frame of the original space $\mathbb{R}^n$
containing $D$, parallel to itself, to the points of $D$. Since $\varphi : D → G$
is a diffeomorphism, the mapping $\varphi′(t) : TD_t → TG_{x=\varphi(t)}$ of tangent
spaces effected by the rule $TD_t \ni \mathbf{e} \to \varphi′(t)\mathbf{e} = \mathbf{\xi} \in TG_x$, is an
isomorphism of the tangent spaces at each point $t$. Hence from the
frame $\mathbf{e}_1,...,\mathbf{e}_n$ in $TD_t$ we obtain a frame $\mathbf{\xi}_1 = \varphi′(t)\mathbf{e}_1,...,\mathbf{\xi}_n = \varphi′(t)\mathbf{e}_n$
in $TG_x$, and the field of frames on $D$ transforms into a field of
frames on $G$ (see Fig. 12.6). Since $\varphi \in C^1(D, G)$, the vector field
$\xi(x) = \xi(\varphi(t)) = \varphi′(t)\mathbf{e}(t)$ is continuous in $G$ if the vector field $\mathbf{e}(t)$
is continuous in $D$. Thus every continuous field of frames
(consisting of n continuous vector fields) transforms under a
diffeomorphism to a continuous field of frames. Now let us consider a
pair of diffeomorphisms $\varphi_i : D_i → G$, $i = 1, 2$, which introduce two
systems of curvilinear coordinates $(t_1^1 ,..., t_1^n )$ and $(t_2^1 ,..., t_2^n )$ into the same domain $G$. The mutually inverse diffeomorphisms
$\varphi_2^{−1} \circ \varphi_1 : D_1 \to D_2$ and $\varphi_1^{−1} \circ \varphi_2 : D_2 \to D_1$ provide mutual transitions
between these coordinate systems. The Jacobians of these mappings at
corresponding points of D_1 and D_2 are mutually inverse to each other
and consequently have the same sign. If the domain $G$ (and together
with it $D_1$ and $D_2$) is connected, then by the continuity and
nonvanishing of the Jacobians under consideration, they have the same sign at all points of the domains $D_1$ and $D_2$ respectively. Hence
the set of all curvilinear coordinate systems introduced in a
connected domain G by this method divide into exactly two
equivalence classes when each class is assigned systems whose mutual
transitions are effected with a positive Jacobian. Such equivalence
classes are called the orientation classes of curvilinear coordinate
systems in G. To define an orientation in $G$ means by definition to fix
an orientation class of its curvilinear coordinate systems.

What he says so far is quite clear, I have only one doubt: why must the two domains necessarily be subsets of $\mathbb{R} ^ n$? Could not $D$ be an open set of $R ^ k$ ($k <n$)? It is true that in this case all the logical path it has taken would be lost because it has used the determinants (therefore square matrices), but an open set of $R ^ k$ is exactly what is used when a chart is defined for a neighborhood of a point of a $k$-dimensional surface in $\mathbb{R}^n$. So it's not a strange practice, No?
Then he continues:

It is not difficult to verify that curvilinear coordinate systems
belonging to the same orientation class generate continuous fields of
frames in $G$ (as described above) that are in the same orientation
class of the tangent space $TG_x$ at each point $x \in G$.

Given that, based on what I've read so far, I don't know what "orientation class of the tangent space $ TG_x $" means, I mean this last sentence like this: if $\varphi_1$ and $\varphi_2$ are in the same class equivalence (i.e. they have the same sign as the determinant of the Jacobian matrix in all points of $D$), then the fields of frames they generate in $G$ are such that in each point $x \in G$ the frame generated by $\varphi_1$ is in the same class as equivalence of the frame generated by $\varphi_2$ (that is, the transition matrix from one frame to another has positive determinant). Is it correct?
Finally, last sentence:

It can be shown in general that, if $G$ is connected, the continuous fields of frames on $G$ divide into exactly two equivalence classes if each class is assigned the fields whose frames belong to the same orientation class of frames of the space $TG_x$ at each point $x \in G$.

I just can't understand this. What does it mean?
 A: 
What he says so far is quite clear, I have only one doubt: why must the two domains necessarily be subsets of $\Bbb{R} ^ n$? Could not $D$ be an open set of $\Bbb R ^ k$ ($k <n$)?

He is talking about changing coordinates. You cannot use a $k$-dimensional coordinate system on an $n$-dimensional space.

If I have a smooth $k$-dimensional surface $S$ in $\Bbb R^n$, then exists a diffeomorphism between a neighborhood of a point $x\in S$ and the unit open cube in $\Bbb R^k$ (with $k\ne n$ in general).

But that is changing $k$-dimensional coordinates on a $k$-dimensional space. He is talking about changing $n$-dimensional coordinates on a domain in $\Bbb R^n$, which is first of all an open subset, and therefore an $n$-dimensional subspace. I'm sure he'll get around to talking about orientations on other spaces later, but at this point he is specifically discussing them for domains.
Have patience.

Given that, based on what I've read so far, I don't know what "orientation class of the tangent space $TG_x$" means

Given two frames, there is a linear map that takes one to the other. That linear map has a determinant which is either positive or negative (if it were $0$, they could not both be frames). Frames related by a map with positive determinant are said to have the same orientation.
In $\Bbb R^2$, two frames are equivalent if rotating $e_1$ to $e_2$ in the shortest direction is a clockwise rotation in both frames, or is a counter-clockwise rotation in both frames. When the plane is imbedded in $\Bbb R^3$, the two frames are equivalent if $e_1 \times e_2$ points to the same side of the plane for both frames.
For frames in $\Bbb R_3$, frames are equivalent if they are both right-handed (with your right hand, you can point the first finger in the diretion of $e_1$, the second finger in the direction of $e_2$ and the thumb in the direction of $v_3$) or if both are left-handed.

I mean this last sentence like this: if $\varphi_1$ and $\varphi_2$ are in the same class equivalence (i.e. they have the same sign as the determinant of the Jacobian matrix in all points of $D$), then the fields of frames they generate in $G$ are such that in each point $x \in G$ the frame generated by $\varphi_1$ is in the same class as equivalence of the frame generated by $\varphi_2$ (that is, the transition matrix from one frame to another has positive determinant). Is it correct?

Since we are working in $\Bbb R^n$, you can get away with that description. But for general manifolds, $d\varphi_1$ and $d\varphi_2$ are maps between two different vector spaces. To define the determinant of a linear map, it has to carry a vector space to itself. So in general, you need to talk about the Jacobian determinant of $\varphi_1^{-1}\circ\varphi_2$ and of $\varphi_2^{-1}\circ\varphi_1$, which are both maps from $\Bbb R^n \to \Bbb R^n$.
Note that now we are no longer talking about the tangent space at a particular point $x$, but rather the tangents spaces over many points. It is conceivable that at some point $x_1, |d(\varphi_1^{-1}\circ\varphi_2)_{x_1}| > 0$ while at some other point $x_2, |d(\varphi_1^{-1}\circ\varphi_2)_{x_2}| < 0$. A key part of this development is that if the domain is connected, this cannot occur. Since the Jacobian determinant is continuous and cannot be $0$ since $\varphi_1^{-1}\circ\varphi_2$ is always invertible, there is no way for it to cross from positive to negative, or vice versa.

I just can't understand this. What does it mean?

This is a more general case of what I just discussed. In that case we were talking about the frame fields (an assignment to each point $x$ of a frame in its tangent space) defined by $\varphi_1$ and $\varphi_2$. Now he is generalizing to any continuous frame fields, not just to those defined by coordinate maps. Given two continuous frame fields $F_1, F_2$, if at some $x_0$, the frame $F_1(x_0)$ has the same orientation as (i.e., is equivalent to) $F_2(x_0)$ and the domain is connected, then by continuity, for every point $x$ in the domain, it must be true that $F_1(x)$ has the same orientation as $F_2(x)$. They cannot agree in their orientations at one point, but disagree at another point. The argument that proves it is very similar to the one above. At each $x$ there is a unique invertible linear map carrying $F_1(x)$ to $F_2(x)$. The determinant of this linear map is a continuous function of $x$, so it cannot switch from positive to negative without passing through $0$, which it cannot do.
Note that the argument breaks down when the domain is not connected. Continuous frame fields can agree on orientation on one component of the domain, and disagree on another component.
