Classically, if $X \subseteq \mathbf{R}^{3}$ is an open set, then a $3$-dimensional coordinate system on $X$ is nothing but an injective, continuously-differentiable mapping $\xi:X \to \mathbf{R}^{3}$ whose differential has rank three at each point. Conceptually, a coordinate system assigns an ordered triple of real numbers to each point of $X$ in such a way that distinct points are given distinct coordinates, satisfying some technical conditions of smoothness. The inverse mapping $\xi^{-1}:\xi(X) \to X$ is sometimes called a parametrization of $X$.
The inclusion map $i:X \hookrightarrow \mathbf{R}^{3}$ is a coordinate system on $X$. If $A$ is an invertible $3 \times 3$ real matrix and $b$ is a vector in $\mathbf{R}^{3}$, then the affine map $T(x) = Ax + b$ defines a coordinate system on $X$. Cylindrical and spherical coordinates are parametrizations of portions of $\mathbf{R}^{3}$; for example, the spherical coordinates mapping (with geographic angles)
$$
S(r, \theta, \phi) = (r\cos\theta\cos\phi, r\sin\theta\cos\phi, r\sin\phi),\qquad
0 < r,\quad |\theta| < \pi,\quad |\phi| < \pi/2
$$
parametrizes $X = \mathbf{R}^{3}\setminus \{y = 0, x \leq 0\}$, the complement of a closed half-plane. The spherical coordinates of a point $(x, y, z)$ of $X$ are the numbers $(r, \theta, \phi)$ such that $(x, y, z) = S(r, \theta, \phi)$.
In linear algebra (over an arbitrary field $F$), a "coordinate system" for an $n$-dimensional vector space $V$ takes values in $F^n$, and is usually defined by choosing a basis $B = \{v_{i}\}_{i=1}^{n}$ and mapping a vector $v$ in $V$ to its coordinate vector $[v]_{B}$ in $F^n$. Similarly, in affine geometry one can construct a coordinate system as you describe (by picking an origin and a basis for the resulting vector space).
In the modern point of view, coordinate systems play a secondary role to "overlap maps". For concreteness, let $X$ be an open subset of $\mathbf{R}^{3}$. First, we weaken the criteria for a mapping to be a coordinate system, requiring only that $\xi:X \to \mathbf{R}^{3}$ be continuous and injective. Now assume $\xi_{1}$ and $\xi_{2}$ are "allowable" coordinate systems on $X$ (for some value of "allowable", yet to be determined). An overlap map is a composition
$$
\xi_{1} \circ \xi_{2}^{-1}:\xi_{2}(X) \to \xi_{1}(X).
$$
The "structure" of $X$ is encoded in the properties of the overlap maps. For example, if the overlap maps are diffeomorphisms, then "smoothness" makes sense for functions $f:X \to \mathbf{R}$: We pick an arbitrary coordinate system $\xi_{1}$ and declare $f$ to be smooth if the composition $f \circ \xi_{1}^{-1}$ is smooth as a function on $\mathbf{R}^{3}$. This definition does not depend on $\xi$, since
$$
f \circ \xi_{2}^{-1} = (f \circ \xi_{1}^{-1}) \circ (\xi_{1} \circ \xi_{2}^{-1}),
$$
and the overlap map is a diffeomorphism. Similarly, if the overlap maps are affine, then (for example) "line segments" in $X$ make sense: A subset $\ell$ of $X$ is a "segment" if in some (hence every) coordinate system $\xi$, the set $\xi(\ell)$ is a line segment.
Philosophically, a coordinate system is merely how one "transfers" objects and functions on a space $X$ to a objects and functions on a "standard" space, such as $\mathbf{R}^{3}$. The interesting "structure" of $X$ is encoded by the overlap maps, which determine properties of $X$ that are independent of the coordinate system.
When mathematicians speak of a manifold having a smooth structure, they mean that some collection of coordinate systems has been fixed so that the overlap maps are diffeomorphisms. A manifold having an affine structure has coordinate systems whose overlaps are affine (a much more stringent condition). Similarly, one hears of holomorphic, piecewise-linear, and conformal structures, among many others.
To address the question about coordinate vector fields in spherical coordinates: Though the spherical coordinates map parametrizes part of $\mathbf{R}^{3}$ (which is a vector space), the spherical coordinates mapping is not a linear transformation, and therefore does not belong to linear algebra, but instead to multivariable calculus.
The "proper framework" requires some explanation. If $X \subseteq \mathbf{R}^{3}$ is an open set, define the tangent bundle $TX$ to be $X \times \mathbf{R}^{3}$. An element $(\mathbf{x}, \mathbf{v})$ of $TX$ should be viewed as consisting of an element $\mathbf{x}$ of $X$ together with a vector $\mathbf{v}$ "based at" $\mathbf{x}$. It makes sense to take linear combinations of vectors only if they're based at the same point.
In this picture, a vector field on $X$ is a mapping $\Xi:X \to TX$ that assigns, to each point $\mathbf{x}$ of $X$, a vector $\Xi(\mathbf{x})$ based at $\mathbf{x}$. The Cartesian coordinate fields $\mathbf{e}_{i}$ are constant vector-valued functions because Cartesian coordinates change by additive constants under translation (!). By contrast, the spherical coordinate fields are non-constant, because spherical coordinates do not change in a simple way under translation.
Best Answer
Say we have a point $P = (r, \theta, z)\in \Bbb R^3$, using cylindrical coordinates. This gives rise to a natural $\vec e_r$ which is of unit length and lies in the $xy$-plane and has angle $\theta$ to the $x$-axis, and a vector $\vec e_z$ which is just the unit vector along the $z$-axis. This way we have $$ P = r\vec e_r + z\vec e_z $$ A different point $Q$ would give a different $\vec e_r$ (and theoretiacally also a different $\vec e_z$, but by a quirk of the cylindrical coordinate system, that vector happens to be the same for any point). Thus $\vec e_r, \vec e_z\in \Bbb R^3$ aren't really well-defined by themselves, but only when given a specific point to work from.
At the same time, at this point $P$, we have the tangent space $T_P\Bbb R^3$. This is also a three-dimensional vector space, like $\Bbb R^3$, but it is not the same $\Bbb R^3$ which $P$ lies in. In this vector space we have a natural basis derived from the coordinate system around $P$, and that basis is sometimes called $\vec e_r, \vec e_\theta$ and $\vec e_z$ (the exact construction of these vectors uses the definition of $T_P\Bbb R^3$ extensively, so it will vary from book to book, and there are also different conventions as to what their lengths should be).
If we have a different point $Q\in \Bbb R^3$, then $T_Q\Bbb R^3$ is also a three-dimensional vector space, but it is different from $\Bbb R^3$ and it is different from $T_P\Bbb R^3$.
A vector ("point") in $\Bbb R^3$ has no natural translation into a vector in $T_P\Bbb R^3$, and neither does a vector in $T_Q\Bbb R^3$. However, both $T_P\Bbb R^3$ and $T_Q\Bbb R^3$ have their own naturally defined basis called $\vec e_r, \vec e_\theta$ and $\vec e_z$, but they only share their name because they are constructed the same way. A vector in $T_P\Bbb R^3$ is completely uncomparable to a vector in $T_Q\Bbb R^3$ (this is most easily seen by just looking at your definition of the tangent space; there is no natural way of making a vector of one space into a vector of the other (well, there is by translation, but that's not applicable to non-flat spaces and this a terrible habit to start)).
Along with the $\vec e_r, \vec e_z$ of $\Bbb R^3$, this is ripe for confusion. This is what I suspect has happened to you, and made you write this question. My best tip is to always be very aware of which vector space any vector is a part of when you write them down and manipulate them. If you do this by actually giving them different names (instead of calling all of them $\vec e_r$), then it's a little bit easier to keep things straight.