Why do displacement vectors not generalise to manifolds? I was reading through the argument as to why displacement vectors, being a vector pointing from one point to another, cannot be generalised to manifolds. It went as follows;

On a manifold, displacement "vectors" aren't very vector-like: because the coordinate transformations aren't in general linear, something that looks like a straight line between two points in one coordinate system isn't going to in another. As a result, we can't define or add displacement vectors in a coordinate-independent way. They don't form a vector space. It only works in at space because we have a preferred set of inertial coordinate systems, which are related by linear coordinate transformations.

I wanted some clarification as to what is intended by the term 'coordinate transformation'. My lecture notes previously defined a coordinate transformation to be a mapping between two coordinate systems over the same set at the point where they overlap. In this case, where the set is a manifold, I can visualise how the argument above holds. For example, in the spacetime around the earth, as you approach the earth, the spacetime becomes more and more curved. Meanwhile, far away from the earth (if we assume in this model that the only gravitational object present is the earth) then the space time is approximately flat. Indeed, near the earth a 'straight line' would be rather curved, while far away from the earth it would be 'straight' as we intuitively think of straight to be.
However, the argument then proceeds to say that it works in flat space as 'we have a preferred set of inertial coordinate systems, which are related by linear transformations'. I would have thought this to be the Lorentz transformation, however this is a transformation between different frames of reference. Therefore, to be consistent with the previous argument I would think that coordinate transformations must refer to transformations between different manifolds, that is between different frames of reference in curved space time.
So in summary I suppose that I'm asking the following; is the coordinate transformation a transformation between coordinate systems of the same manifold, or is it a transformation between coordinate systems of two different manifolds entirely? Any explanations so as to correct the flaws of my understanding, or to expound on the reasoning as to why displacement vectors cannot be generalised to manifolds, would be much appreciated. Thanks :)
 A: It is helpful to think of how a manifold is defined formally, to understand what a coordinate transformation is.
One can think of smooth manifolds as follows. A manifold is a locally Euclidean space, meaning that at any given point $p$, there is some open set $U\subset M$ containing $p$, where $U$ is homeomorphic to an open set $\tilde{U}\subset \mathbb{R}^n$. The homeomorphism $\phi: U\to \tilde{U}$ is called a coordinate chart. When we speak of a coordinate system we are referring to this function $\phi$. So an element $p\in U$ has is said to have the coordinates $(p^1, p^2,\cdots, p^n)=\phi(p)\in \tilde{U}$.
Since in special relativity we deal with smooth manifolds and objects that vary smoothly over space-time, we need a way to talk about derivatives and smooth maps. To do this we require overlapping coordinate patches to have differentiable transition functions, that is, given $p\in M$ and $U, V$ two coordinate patches containing $p$ with charts $\phi,\psi$ respectively, the function $\psi\circ \phi^{-1}$, which is a function from an open set in $\mathbb{R}^n$ to $\mathbb{R}^n$, to be infinitely differentiable.
If we have a function $f: M\to \mathbb{R}$, it has a representative function $\tilde{f}=f\circ \phi^{-1}$, which gives $f$ as a function of the specific coordinates of $\phi$. If we didn't require the transition functions to be differentiable, we would quickly run into an issue. We might have a function $\tilde{f}$ which is differentiable in our original coordinate system, but when we switch to some coordinates given by $\psi$, then $\bar{f}=f\circ \psi^{-1}=f\circ \phi^{-1}\circ \phi\circ \psi^{-1}$ could end up not being smooth. The differentiability of the transition functions means that when we change coordinates, the representation of functions, tensors, and whatnot remain smooth.
One can see the issue of displacement vectors even $\mathbb{R}^3$. Taking the standard euclidean coordinates on $\mathbb{R}^3$, $\phi: \mathbb{R}^3\to \mathbb{R}^3$ is just the identity. We can speak of the displacement vector between two points $p=(p^1,p^2,p^3)$ and $q=(q^1,q^2,q^3)$, this is given by $p-q=(p^1-q^1, p^2-q^2,p^3-q^3)$. If we wanted to go to cylindrical coordinates, we could map $p-q$ by our new coordinate transformation. When we're in cylindrical coordinates, we could feasibly talk about displacement vectors here too. if $p=(r,\theta,\varphi), q=(\rho, \vartheta, \psi)$ we could just as easily say that the displacement vector between these two points is $p-q=(r-\rho, \theta-\vartheta, \phi-\psi)$, but since the transformation between our first coordinates and cylindrical coordinates is not linear, we will have trouble getting a consistent idea of what a displacement vector even is. In this case, it can be fixed by just always calculating the displacement vector in the natural coordinates and transform it from there into the other coordinates, but on a manifold where there is no natural coordinate system to declare the correct one, we have no consistent way of doing this.
