First of all, I'm not sure if this question even makes sense, i.e. is there a notion of curvature on a vector space structure. However, when dealing with vector spaces (here I am mostly thinking of tangent spaces on a manifold), we treat (and draw) them as flat. On the other hand, I don't see the flatness assumption (nor any kind of parallelism) in the vector space axioms.
So, my question is, are and what makes vector spaces flat? What is the difference between the arrows in flat space and e.g. directed arcs of great circles on a 2-sphere? Don't they both satisfy the same axioms?
I think it's not about curvature, there is something more fundamental that distinguishes $\mathbb{R}^n$ from any other manifold. For any $p\in\mathbb{R}^n$ there is a canonical isomorphism $T_p\mathbb{R}^n\to\mathbb{R}^n$. Any tangent vector $v\in T_p\mathbb{R}^n$ is represented by some path $\alpha$ with $\alpha(0)=p$. The isomorphism is given by $v\mapsto\dot{\alpha}(0)\in\mathbb{R}^n$.
This isomorphism yields a canonical isomorphism between the tangent spaces at two distinct points, and this is something that does not exist on any other manifold.
The concept of parallel transport is thus very simple on $\mathbb{R}^n$. Actually, a vector field on $\mathbb{R}^n$ can be constant, where the definition of constant is very clear.
Intuitively speaking, curvature of a Riemannian manifold $M$ measures how much two vector fields don't commute. But in Euclidean space, having noticed that any tangent vector can be extended canonically to a constant vector field, the fact that every two tangent vectors commute is just the theorem of changing the order of differentiation:
$$\frac{\partial^2}{\partial x\partial y}=\frac{\partial^2 f}{\partial y \partial x}.$$
Since every two tangent vectors (or vector fields) commute, the curvature is $0$.