differential geometry - How can a set of components fail to make up a vector?

Many books in Physics insist to define vectors are objects with components with the property that the components transform in a proper way under a change of coordinates. Now, in mathematics, on the other hand, vectors as geometrical objects (rather than the algebraic objects from linear algebra) belong to the realm of Differential Geometry. In that case, we have a smooth manifold $M$ and a point $a\in M$. A vector at $a$ can be defined:

1. As an equivalence class of smooth curves passing through $a$, which intuitively, at that point are going in the same direction.


2. As a point derivation, that is, a derivation on the algebra of germs of smooth functions at $a$.

This defines the tangent space $T_a M$. This is a real vector space with dimension $n = \dim M$, and hence $T_a M \simeq \mathbb{R}^{n}$. In particular this means there is a bijection between $T_a M$ and $\mathbb{R}^n$ so that given any tuple of components, they do form a vector on $T_a M$. This seems to be against the physicists' definition, since there's nothing imposed on the components to form a vector.

On the other hand, we can put all tangent spaces together to form the tangent bundle $TM$. We define then vector fields as sections of that bundle, that is, mappings $X: M\to TM$ such that $\pi\circ X = \operatorname{id}_M$ where $\pi : TM\to M$ is the natural projection. $X$ should also be a continuous and differentiable, and of course, it also has components given by

$$X = X^i \dfrac{\partial}{\partial x^i}$$

Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go.

So my questions are:

1. What is really the point in giving so much enphasis on the way vectors transform, coming to the point that we use this property to even define vectors?



2. 
   

   When physicists talk about defining a vector using transformation properties, are they really talking about vector fields and changes of coordinates on a manifold or vectors and changes of basis on each tangent space?

   
   


3. 
   

   How can a set of components (or component functions) can fail to make up a vector (or vector fields)?

   
   

Answer

OP wrote (v3):

Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go.

It is (implicitly) implied by OP's notation that

1. 
   

   the component functions $X^1,\dots, X^n\in C^{\infty}(M)$ are globally defined functions.

   
   


2. 
   

   the coordinate functions $x^1,\dots, x^n\in C^{\infty}(M)$ are globally defined functions.

   
   

However, there are many examples of differentiable manifolds $M$, that don't have a global coordinate chart, e.g. the 2-sphere $S^2$.

The general notion of a vector field should not rely on whether there exists a global coordinate chart, or not.