I'm confused as to how orientations and ordered bases relate to the “handedness” of Cartesian coordinates. On page 237 of Tu's An Introduction to Manifolds he's discussing orientations of a vector space and writes, with reference to the following diagram:
On $\mathbb{R}^{3}$ an orientation is either right-handed (Figure 21.3) or left-handed (Figure 21.4). The right-handed orientation of $\mathbb{R}^{3}$ is the choice of a Cartesian coordinate system such that if you hold out your right hand with the index finger curling from the vector $e_{1}$ in the $x$-axis to the vector $e_{2}$ in the $y$-axis, then your thumb points in the direction of of the vector $e_{3}$ in the $z$-axis.
On my reading, “the choice of a Cartesian coordinate system” implies that Figure 21.4 should therefore show a left-handed Cartesian coordinate system. But that doesn't make sense as then $e_{2}$ would lie on the $x$-axis and $e_{1}$ would lie on the $y$-axis. On reflection (excuse pun) I'm guessing that an ordered basis defines an orientation on $\mathbb{R}^{3}$, which is then graphically represented by choosing either a right-handed or left-handed Cartesian system. So both Tu's diagrams refer to a right-handed coordinate system. Am I on the right track?
EDIT In other words, how do Tu's diagrams relate to these right and left-handed coordinate systems?
EDIT 2 Part of my confusion stems from this answer by @AndrewD.Hwang here:
It's worth pointing out a final subtlety in $\mathbb{R}^{3}$: If $\left(\mathbf{e}^{i}\right)_{i=1}^{3}$ denotes the standard basis, the (algebraic) ordering of the vectors fixes an orientation, but we must still pick a conventional geometric representation, i.e., must choose to depict the standard basis as "right-handed" or "left-handed".
On p. 238 Tu gives a formal definiton of orientation as an equivalence class of orderded bases. In the sense of this definiton any (finite-dimensional) vector space $V \ne \{0\}$ has exactly two orientations.
Note that $V = \{0\}$ is an exception. Formally it has $\emptyset$ as its only base (if we agree that sums over the empty index set are $0$, then each element of $\{0\}$ has a unique representation as a linear combination of basis vectors) and thus it has only one orientation.
Tu's preceding explanations are purely motivational and may in fact be somewhat confusing. What is he doing?
In dimensions $n = 1, 2, 3$ he considers the standard Cartesian coordinate system in $\mathbb R^n$ and assigns two orientations to it. Be aware that such an orientation is an additional component attributed to the coordinate system, it cannot be found in the coordinate system itself. It occurs in the figures in form of rotation arrows which are assigned in two variants to the depicted Cartesian coordinate system.
Let us focus to $n = 3$. Tu considers the usual $x$-$y$-$z$-coordinate system with standard unit vectors $e_1$ on the $x$-axis, $e_2$ on the $y$-axis and $e_3$ on the $z$-axis. Drawing the $e_i$ as arrows gives a direction to each of the axes (the positive direction). From the viewer's perspective the graphical representation of Cartesian coordinates in 21.3 and 21.4 shows the $x$-axis pointing forward, the $y$-axis to the right and the $z$-axis upwards. This is an arbitrary convention, you could do it as you want. As an example take the figure in your edit. Anyway, if you decided for a particular graphical representation, you can glue the coordinate system to your right hand or your left hand according to the following convention:
Arrange index finger (I), middle finger (M) and thumb (T) so that they are perpendicular to each other. Then the fingers of your right and of your left hand form two ordered $3$-frames, the right-handed frame $(I_R, M_R, T_R)$ and the left-handed frame $(I_L, M_L, T_L)$. No spatial movement of your hands will transform the right-handed into the left-handed frame - they are mirror images. These two frames are the intuitive prototypes of an orientation of $\mathbb R^3$.
Gluing to the right hand ("right-handed orientation"): Point with $I_R$ in direction $e_1$, with $M_R$ in direction $e_2$ and with $T_T$ in direction $e_3$.
Gluing to the left hand ("left-handed orientation"): Point with $I_L$ in direction $e_2$, with $M_L$ in direction $e_1$ and with $T_L$ in direction $e_3$.
This attributes a handedness or orientation to the Cartesian coordinate system. Formally this is described via the two bijections $$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_2 & e_1 & e_3 \end{pmatrix}$$
In my opinion the graphical representation in your edit is more intuitive than Tu's.
In both cases you use assign $e_1$ to $I$, $e_2$ to $M$ and $e_3$ to $T$ which gives you two variants of a Cartesian coordinate system which are optically distinguishable. Here you have $$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & e_2 & e_3 \end{pmatrix}$$
However, if you take the graphical representation on the right side as the standard representation of a Cartesian coordinate system (this agrees with Tu, just imagine to look at this coordinate system with your eye facing the arrow tip of $x$), then you will admit that the graphical representation on the left side is non-standard. And in fact, if you use your left hand in the standard representation, then you get $$\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & -e_2 & e_3 \end{pmatrix}$$ Working with equivalence classes of ordered bases, you can easily see that the latter representation of $\omega_L$ coincides with Tu's variant.