This is not a proper answer, but it might be useful anyway. It's just what I've found so far reading books and trying to make sense of everything I learned about vectors in both calculus and algebra.
You can geometrically picture vectors in $\mathbb{R}^n$ as arrows placed at the origin. Every vector can be uniquely expressed as a linear combination of $n$ linearly independent vectors, so for every basis of this vector space $(\vec{e}_1,\dots,\vec{e}_n)$ we can write:
$$
v=v^1 \vec{e}_1 + \cdots + v^n \vec{e}_n
$$
Although using a vector basis allows us to uniquely identify every vector in $\mathbb{R}^n$, it isn't really useful when trying to identify every "point" in $\mathbb{R}^n$ because of its limitations (your axes will have to be straight lines and will have to include the "canonical" origin).
In order to identify every point of $\mathbb{R}^n$ with more freedom, our first approach could be affine geometry. In $\mathbb{R}^n$ viewed as an affine space, we can define a coordinate system $(O,\mathcal{B})$, with $O$ a point in $\mathbb{R}^n$ and $\mathcal{B}$ a basis of $\mathbb{R}^n$ as a vector space. Axes are still straight lines, but now we can move from the origin to any other point $O$. This affine coordinate system is in a way a coordinate system, and it is definitely not the same as a basis — because $(O,\mathcal{B}) \neq \mathcal{B}$ — but we can do better.
We can define a system of $n$ equations (not necessarily linear, a system of linear equations would bring us back to the affine case) that uniquely identify every point in $\mathbb{R}^n$:
$$
x^i =\Phi^i(q^1,\dots,q^n), \space\space\space i=1,\dots,n
$$
This is precisely what we do when we define cylindrical or spherical coordinates: express $x,y,z$ in terms of three new variables ($\rho,\varphi,z$ and $r,\theta,\varphi$ respectively).
This means that the values $(q^1,\dots,q^n)$ will be our new coordinates. Note that, if this system were linear, we would need to require the coefficient matrix to have a non-zero determinant in order for this system to have a (unique) solution. For a general system of equations, by virtue of the implicit function theorem, the analogous condition is:
$$
\frac{\partial(x^1,\dots,x^n)}{\partial(q^1,\dots,q^n)} \neq 0
$$
i.e. the Jacobian of the transformation must be non-zero.
This new coordinates $(q^1,\dots,q^n)$ aren't related to any basis, but they induce one for every point $(q^1,\dots,q^n)$ in $\mathbb{R}^n$: the so-called coordinate basis of this coordinate system:
$$
\vec{v}_\mu = \frac{\partial\vec{\Phi}}{\partial q^\mu}
$$
In this sense, we can see than the components of the position vector at the point $p\in\mathbb{R}^n$ will be different from the coordinates of the point $p$ itself. The components depend on the coordinate basis (or any other basis which you define in terms of that one), while the coordinates of a point depend on the coordinate system itself.
In fact, now we're not talking about $\mathbb{R}^n$ as a vector space anymore, but this space does have a vector space attatched at every point, a vector space we call the tangent space at $p$: $T_p\mathbb{R}^n$. The coordinate basis at every point is the vector basis that our coordinate system induces for the tangent space at that point.
Note that the components of a vector (or a tensor, for that matter) may be called coordinates. I only see this when reading about pure vector spaces, without any sense of geometry, metrics or anything. For example, a matrix $A\in\mathcal{M}_{n\times n}$ which satisfies $\vec{v}_{\mathcal{B}'} = A\vec{v}_{\mathcal{B}}$ might be called a change of coordinates matrix (from the coordinates from $\mathcal{B}$ to the coordinates from $\mathcal{B}'$) or a change of basis matrix (from the basis $\mathcal{B}'$ to the basis $\mathcal{B}$, because it satisfies $\mathcal{B}'A=\mathcal{B}$, if we allow ourselves this abuse of notation).
Nonetheless, I always say components when referring to a vector (or tensor), and coordinates when referring to a point.
A vector is an element of a vector space. An element of a vector space can be an $n$-tuple of numbers, a polynomial, a matrix, a function etc.
A linear transformation transforms a vector ($n$-tuple, polynomial, matrix, function, etc.) into another vector ($n$-tuple, polynomial, matrix, function, etc.). A matrix cannot transform a vector into another vector, because you can multiply a matrix by an $n$-tuple, but you can't multiply a matrix by a polynomial, a matrix (well, not always, see below), a function, etc.
A matrix associated to a linear transformation can only multiply $n$-tuples of coordinates respect to a basis, and the results are $n$-tuples of coordinates respect to a basis.
Imagine that your vector space is the set of all symmetric $2\times 2$ matrices, and that your linear transformation is:$$T\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\begin{bmatrix} c & a \\ a & b \end{bmatrix}$$
The simplest basis is: $\left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right\}$.
Respect to this basis the coordinates of $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$ are $(a,b,c)$, the coordinates of $\begin{bmatrix} c & a \\ a & b \end{bmatrix}$ are $(c,a,b)$.
The matrix associated to $T$ respect to that basis is: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$.
You can't multiply $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ by $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$, but:
$$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} c \\ a \\ b \end{bmatrix}$$
i.e.
$$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\text{Coord}\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\text{Coord}\left(\begin{bmatrix} c & a \\ a & b \end{bmatrix}\right)$$
This is why:
- you always need a basis to associate a matrix to a linear
transformation (when the basis is omitted you assume the canonical basis),
- the matrix associated to a linear transformation is unique respect to
a fixed basis,
- you can also have different bases for the domain and the range of a
linear transformation, so the matrix associated to a linear transformation is unique respect to the basis of its domain and the basis of its range,
- since there are infinite bases, there also are infinite matrices
associated to a linear transformation.
Best Answer
In a nutshell, yes, you might be thinking it just a tad wrong.
The thing is, when you work in terms of linear algebra, you basically work within fixed vector spaces. That is, whenever you start a new chapter, that'll most likely include: "consider a vector space $V$, with the dimensionality of $n$ and a scalar product given by the rule $\varphi: V\times V \rightarrow \mathbb{R}$". And here, explicitly, you know that you have got a vector structure, hence you've got a basis, hence you can expand any vector within this vector space as a linear combination of basis vectors.
On the other hand, when you deal with differential geometry, especially with manifolds, you never get a vector structure in the first place and that's why manifolds matter. The sphere $\mathbb{S}^2$, as a manifold, exists regardless of the Euclidean space it can be embedded into.
Imagine this: an eternal creature creates a sphere and you on that sphere. Apart from anything, you don't know what it is to have a vector on a manifold, that is just outside of the scope of your comprehension. Now the first thing you do, is define the coordinates, i. e., you state the rules of assigning each point $P \in \mathbb{S}^2$ a specific set of numbers $(u_1^P, u_2^P, \dots, u_n^P)$. Once you have defined such numbers $\forall P$ in some neighbourhood, and those numbers change continuously from a point to a point, then you have defined the local coordinates.
As it happens on manifolds, prior to introducing the notion of a vector, you have to introduce the notion of coordinates. Once you have the latter, you unite the sequence of points in a path and get a curve on the manifold. That curve $\gamma(t): [a, b]\rightarrow M$ is a function of coordinates: $$\gamma(t) = [u_1(t), u_2(t), \dots, u_n(t)].$$
The rest is simple: you define a tangent vector $\xi$ at a point $P=\gamma(t_0)$ as a velocity vector of the curve: $\xi = \frac{d}{dt}\gamma(t)\big|_{t_0}$. It is only now that you have created a connection between a manifold and some sort of a vector structure.
After that you prove that $\partial_{u_1}, \partial_{u_2}, \dots, \partial_{u_n}$ form a basis in $T_{P}M$ and conduct all the analysis with tangent spaces / bundles and such.