Do arrows on the surface of a sphere also form a vector space? Given 2 points $a$ and $b$ on the plane $\mathbb{R}^2$, one can "go" from $a$ to $b$ by drawing an "arrow" from $a$ to $b$. These "arrows" have the structure of an additive group, and they can also be "scaled" by elements of $\mathbb{R}$.
Given 2 points $a$ and $b$ on (say) the 2-sphere, one can "go" from $a$ to $b$ by drawing a "geodesic arrow" from $a$ to $b$. Do such arrows on spheres (and maybe other manifolds) also form groups, or even vector spaces?
 A: $\newcommand{\Reals}{\mathbf{R}}$For each point $a$ of the sphere, the set of arrows based at $a$ and tangent to the sphere at $a$ forms a real vector space, the tangent space $T_{a}S^{2}$.
The collection of all tangent spaces is not a vector space, but a vector bundle, the tangent bundle $TS^{2}$ of the sphere.
(A continuous vector field on the sphere is a continuous "section" of $TS^{2}$; the set of all continuous vector fields is a real vector space, as well as a module over the ring of continuous functions, loosely, a vector space where scalars are continuous, real-valued functions on the sphere.)
Assuming I understand your verbal description, if $a$ and $b$ are distinct, the "arrow from $a$ to $b$" describes a chord of the sphere. The sum of two such arrows only rarely determines a chord, and scalar multiplication determines a chord if and only if the scalar is $0$ or $1$: If you translate the sphere so that $a$ sits at the origin of $\Reals^{3}$, the set of chords is the set of points on the sphere regarded as displacement vectors from the origin, which is obviously not a vector subspace of $\Reals^{3}$.
(Judging from your comment, it's possible my chord interpretation is not what you have in mind. Note, however, that assigning two coordinates to points on the sphere and then using these coordinates as vector components certainly does not turn the sphere into a vector space in any natural way.)
With "geodesic arrows", it appears you're seeing the structure on the tangent bundle mapped to the round sphere by the exponential map. This scheme does not turn the sphere into a vector space. For example, if $a$ and $b$ are antipodal, then (with the obvious definition of addition as "sliding and concatenation") two times any geodesic arc from $a$ to $b$ is the zero arc from $a$ to $a$.
Similar remarks hold for other smooth manifolds.
A: If vectors are considered to be translations (translating the whole space) operating on points then they form a commutative transformation group as illustrated in the figure below:

The translations first $NE$ and then $OS$ move the point $X$ to a new location. It does not matter that we moved the points $OS$ by applying $NE$ first. The translations first $OS$ and then $NE$ moves $X$ to the same place and it does not matter that applying $OS$ moved the points $N$ and $E$ to a new location first.

This is not the case when our points are on a sphere. A pair of points can be considered as a translation of the whole space again. But this translation group will not be commutative. Look at the following figure.

We have a sphere and two pairs of points $NE$ and $OS$. Both pairs translate (rotate) the sphere. Let's see the effect of these translations if applied in different order. 
First, apply $NE$. This translation will move $X$ to the north pole. But, also, this translation will move $S$ to the original place of $X$. If we apply this new $OS$ then $X$ will stay.
However, if we apply $OS$ first then $X$ will stay where it was originally. At the same time $N$ will be moved to the original location of $O$ and $E$ will stay. The new translation $NE$ will move $X$ to the original location of $O$.
So, there is a fundamental difference between the translation group of vectors and the translation group of rotations.
