What does affinity means? (In categorical terms) I know that traditionally, an affine space is "what is left from a vector space, after removing the origin". Given any set $X$ and a ring $K$, we can consider the set of all formal affine linear combinations of X (affine here means the coefficients must add up to 1), as an affine space. (So there is a Set $\stackrel{F\left(X\right)} \longrightarrow$$ AffineLeftModules functor. What does the fact that the coefficients should add up to 1 has to do anything with the intuition about affine spaces? More generally, What can be said in general about a category, whose objects have "affine spaces"-like properties, in a meaningful way?
 A: An affine space over a field $k$ is a set equipped with a family of operations called affine linear combination, one for each tuple $c_1, ... c_n \in k$ that adds up to $1$, which is supposed to behave like the operation $(v_1, ... v_n) \mapsto c_1 v_1 + ... + c_n v_n$ does on a vector space. These operations satisfy various compatibility relations which guarantee that affine linear combinations compose the way they ought to. You can think of this as an object with less structure than a vector space in that in a vector space you would allow arbitrary tuples. More precisely, there is an obvious forgetful functor from vector spaces to affine spaces and you can think of the data it forgets as the origin. 
The category of affine spaces is closely related to the category of heaps. It admits a forgetful functor to $\text{Set}$, and the functor you describe is its left adjoint. 
The geometric intuition is that you can't add elements of an affine space, but you can take generalized averages; for example, over $\mathbb{R}$ you can find midpoints. See also convex space.  
The algebraic intuition is that you start with a vector space but restrict yourself to taking as operations not all linear combinations, but only those which are covariant under translation (that is, under moving the origin) in the sense that 
$$\text{operation}(v_1 + v, ... v_n + v) = \text{operation}(v_1, ... v_n) + v.$$
These are precisely the ones whose coefficients sum to $1$. 
