Difference between "frame" and "coordinate system" From Zorich, Mathematical Analysis II, 1st ed., pag.170-171:

that up to now is sufficiently clear. Then:

So, some questions...
What is a "coordinate system" in strict sense? And what is the difference between it and the "frame" in $\mathbb{R}^n$?
I could think that a frame is a fixed basis $\{\mathbf{e}_1,...,\mathbf{e}_n\}$ and a coordinate system is a function $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ that maps $(t_1,...,t_n)$ into $t_1\mathbf{e}_1+...+t_n\mathbf{e}_n$. To be honest, however, I'm not so convinced, maybe I'm missing something big.
The second question is on understanding something more about the matrix relationship that the author mentions (the transposition of the matrix), but I think it's a simple consequence of what I didn't understand about the first question.
In any case, it would be very useful if someone could help me understand "visually" these two equivalent approaches that Zorich speaks of.
 A: You're not missing anything big, it's just some simple linear algebra. It might help to get hold of an abstract linear algebra textbook. (I won't comment though on your question about visual representation, that's another issue, perhaps suited to a different question).
A frame is an ordered basis $(\mathbf e_1,...,\mathbf e_n)$ of $\mathbb R^n$.
A coordinate system can be defined a bit more abstractly: it is a linear isomorphism $T : \mathbb R^n \to \mathbb R^n$. You may also think of a coordinate system as an ordered basis $(t_1,...,t_n)$ of the dual space $\text{Hom}(\mathbb R^n,\mathbb R)$. The relation between the two points of view of a coordinate system is that $T$ corresponds to $(t_1,...,t_n)$ if and only if $T(x) = (t_1(x),...,t_n(x))$ for all $x \in \mathbb R^n$. This relation is a bijection (between the set of linear isomorphisms of $\mathbb R^n$ and the set of ordered basis of the dual space).
The relation between frames and coordinate systems is that $(\mathbf e_1,...,\mathbf e_n)$ corresponds to $(t_1,...,t_n)$ if and only if for all $x \in \mathbb R^n$ we have
$$x = t_1(x) \, \mathbf e_1 + ... + t_n(x) \, \mathbf e_n
$$
This relation is also a bijection (between the set of ordered bases of $\mathbb R^n$ and the set of ordered basis of the dual space $\text{Hom}(\mathbb R^n,\mathbb R)$.
One last comment regarding your second question, which as you say should be understandandable now, but again it has an abstract formulation. Suppose you the standard basis for $\mathbb R^n$. You then have a corresponding standard basis for the dual space $\text{Hom}(\mathbb R^n,\mathbb R)$, as said above. Each linear transformation $T : \mathbb R^n \to \mathbb R^n$ is then represented by a matrix $M$, as I'm sure you know. But then, there is an associated dual linear transformation $T^* : \text{Hom}(\mathbb R^n,\mathbb R) \to \text{Hom}(\mathbb R^n,\mathbb R)$, defined by $T^*(t)(x)= t(T(x))$, and the key fact is that the matrix of $T^*$ is equal to the transpose of the matrix of $T$.
