What is the name for a vector space on a flat torus? One could define coordintes $(x,y)$ on a torus such that:
$$(x+r_1,y+r_2) = (x,y)$$
Thus one could also do vector operations in these coordinates such as adding. (Except the length of a vector would be ambiguous although it would have a minimum value.)
Is there a name for a vector space on a flat closed manifold such as this?
Or does the term vector space only apply to infinite spaces like $\mathbb{R}^n$ ?
 A: I think the conceptual difficulty here is that you invoke properties of your space as a manifold, which coincides locally with a vector space
(or at least, the tangent space at some point on the manifold is a vector space).
But a manifold is not necessarily globally a vector space.
If we think of the torus embedded in $\mathbb R^3$ in the usual way then the tangent spaces generally do not extend to vector spaces over the whole manifold.
So I will to try to construct a space like yours without reference to the words "torus" or "manifold."
For our space, set two fixed positive real numbers $a$ and $b$; 
then take $\mathbb R^2$ and identify $(x+a,y)$ and $(x,y+b)$ with $(x,y)$
for all real $x$ and $y.$
Let a "vector" be any equivalence class under that identification of points.
Write $[x,y]$ as shorthand for $\{(x+ma,y+nb)\mid m\in\mathbb Z, n\in\mathbb Z\}.$
Define an addition operator $[x,y]+[x',y'] = [x+x',y+y']$ and prove that this is definition actually makes sense (in particular, you get the same result no matter which representative elements $x,$ $y,$ $x'$, and $y'$ you use to describe each of the equivalence classes).
Take $\mathbb R$ as the scalar field.
The addition axioms work fine: 
$\newcommand{X}{\mathbf X}\newcommand{Y}{\mathbf Y}
\newcommand{Z}{\mathbf Z}\newcommand{z}{\mathbf 0}
\X+\Y=\Y+\X,$ $(\X+\Y)+\Z=\X+(\Y+\Z),$ $\z+\X=\X+\z=\X$ where
$\z = [0,0],$ and $\X+(-\X)=(-\X)X+\X=\z$ where $-[x,y]=[-x,-y].$
But how do we define scalar multiplication?
Consider $\frac12 \left[\frac a2, 0\right].$
Note that $\left(\frac a2,0\right) \in \left[\frac a2, 0\right]$
and also $\left(\frac {3a}2,0\right) \in \left[\frac a2, 0\right]$.
So if we implement $r [x,y]$ in the obvious way,
$r \X \stackrel?= [rx,ry]$ for any $(x,y) \in \X,$
we find that we want 
$\frac12 \left[\frac a2, 0\right] \stackrel?= \left[\frac a4, 0\right]$
but also $\frac12 \left[\frac a2, 0\right] \stackrel?= \left[\frac {3a}4, 0\right],$
which implies $\left[\frac a4, 0\right] \stackrel?= \left[\frac {3a}4, 0\right].$
Also note that $(a,b)\in \z,$ so $\left(\frac a2,\frac b2\right)\in \frac12 \z,$
which leads to $\left[\frac a2,\frac b2\right] \stackrel?= \z.$
Proceeding in this way, the set of all vectors with rational coordinates collapses onto the zero vector.
We could rescue things by declaring that in addition to the equivalences already given, every vector is $[0,0],$ but that makes for an uninteresting vector space.
I don't see a way to make this work the way I think you would like it to.
I think you could construct a module on the "flat torus" by taking the vectors defined above along with the ring of integers as your scalar ring.
Then multiplication really is just like repeated addition, and I think you will have a reasonable structure.
