Tangent spaces at different points and the concept of connection If $M$ is a smooth manifold and $TM$ is the tangent bundle clearly $T_pM\cong T_qM$ (as vector spaces) for every $p,q\in M$. Nobody ensures that the previous vector spaces isomorphism is natural (or canonical). In $\mathbb R^n$ we have that $T_p\mathbb R^n$ and $T_q\mathbb R^n$ are naturally isomorphic to $\mathbb R^n$ so we can differentiate a vector field along a direction, in the usual way so taking the directional derivatives of each component.
If the isomorphism between tangent spaces in different point isn't natural, why can't we differentiate a vector field in the usual way? The problem is comparing vectors belonging in different (isomorphic) vector spaces; but we can send the two vectors, with an isomorphism, in a common vector space and then subtract them. Where is the importance of a natural isomorphism?
 A: If a manifold has curvature, then parallel transport of vectors depends on the path. In other words, any map between tangent spaces at different points is dependent on an arbitrary choice which is particular for each pair of points and each particular manifold. It is the same problem as when relating a finite-dimensional vector space $V$ with its dual $V^*$. In order to do so, we need to fix a basis for $V$.
In regards to the importance of naturality, informally a transformation is called natural if it is "independent of its source" in a sense. For example, the isomorphism between $V$ and $V^{**}$, the double algebraic dual, is natural. We send $v\in V$ to $v^{**}\in V^{**}$ such that $v^{**}(f)=f(v)$ for all $f\in V^*$. Note that this is independent not only of the basis of $V$, but also does not reference anything particular about $V$, such as a basis, and so this isomorphism can be carried out consistently over all finite-dimensional vector spaces over a common field. Quoting Wikipedia, a transformation is not natural if it cannot be extended consistently over the entire category in question.
A: Why don't you let $M = S^2 \subset \mathbb{R}^3$ ?   The equation for the Tangent plane is
$$T_{(a,b,c)}S^2 = \{ (x-a)a+(y-b)b+(c-z)c = 0 \}$$
This is a collection of planes parameterized by points $(a,b,c): a^2 + b^2 + c^2 = 1$.  
We can define an initial vector $\vec{x}=(x,y,z) \in T_{(a,b,c)}S^2$.  Then we can ask $$\vec{x}(t + \delta t) = A(t) \vec{x}(t) \in T_{(a,b,c) + \vec{v}\delta t}S^2$$
For any time-paramterized matrix $A$, we should get a connection on the sphere that necessarily Levi-Civita connection, since this flow is not the geodesic flow.
