Why $\mathbf{R}^{n}$, instead of "abstract flat spaces"? An undergraduate taking a course in modern algebra is repeatedly reminded of the power of thinking "abstractly", that is, working with axioms instead of focusing one's attention solely on concrete models. Examples: abstract vector spaces instead of $\mathbf{R}^{n}$, abstract groups instead of group of symmetries or groups of permutations, abstract rings instead of $\mathbf{Z}$, and so on. 
However, modern geometry isn't like that, at least according to the undergraduate level books I've used. Instead of talking about "abstract Euclidean geometry", geometers work with $\mathbf{R}^{n}$ equipped with the usual Euclidean structure, which is a very specific model. Smooth manifolds are built out of patches of $\mathbf{R}^{n}$. Geometers (I think?) don't talk about "the abstract Euclidean plane", they use $\mathbf{R}^{2}$. The sphere is defined as a particular subset of $\mathbf{R}^{3}$, and so on. It would seem reasonable to axiomatize the notion of a "flat space", and then using this to define manifolds, but geometers directly use $\mathbf{R}^{n}$, with the implicit assumption that the model $\mathbf{R}^{n}$ has all the properties of "flatness" that match our intuition. I suppose one could pretend that there exists an axiomatization of flatness (Euclid's axioms), with $\mathbf{R}^{n}$ only a model of the axiomatization, but the viewpoint of modern geometry is that Euclid's axioms are mainly of historical interest. This opinion holds for more rigorous axiom systems (e.g. Hilbert's) as well.
With that being said, I can now turn to the main question: why $\mathbf{R}^{n}$, and not "abstract flat spaces"?
As a sidenote, there's an expository article by Gowers which left me more confused, which can be viewed here. He claims that the set of all $\mathbf{v} \in \mathbf{R}^{3}$, with $\|v\| =1$, is only a model of the sphere, and not the sphere itself. This is puzzling, because all the books I've read define $\mathbb{S}^{2} = \{(x,y,z) \in \mathbf{R}^{3} : x^{2}+y^{2}+z^{2} = 1 \}$. What is an "abstract sphere", then?
 A: Category theory could satisfy you : quickly speaking, you will say that two elements are the same if there exist a "nice identification" between the two. The term "nice" depends of course of the context : in group theory, two groups are the same if they are isomorphic, whereas two topological spaces will be the same if they are homeomorphic... 
To deal with the two examples you are giving in the question, $\mathbb{R}^n$ is not the "perfect abstract vector space of dim $n$", but a result from the course tells you that if you take any vector space $V$ of dim $n$, then it will be isomorphic (in the sense there is a linear bijective map between $V$ and $\mathbb{R}^n$) to $\mathbb{R}^n$, so $\mathbb{R}^n$ will always be a representative of the equivalence class of vector spaces of dim $n$ in the category of vector space. It is the same for the sphere : $\mathbb{S}^2\subset\mathbb{R}^3$ will just be a representative of the sphere's equivalence class for diffeomorphic (if you are in the category of differentiable manifolds) or homeomorphic (topological manifolds) identifications, but also are other models of sphere such as $\mathbb{S}^1=\{x^2+y^2=1\,;\,(x,y)\in\mathbb{R}^2\}\simeq (\mathbb{S}^1)'=\{|z|=1\,;\,z\in\mathbb{C}\}$...
A: First of all, one has to decide what does "geometry" mean in the context of this question, as well as the meaning of a "flat space". 


*

*Geometry might actually mean "topology of manifolds" (possibly differential topology). One typically introduces, say, topological, manifolds, by requiring them to be locally homeomorphic to ${\mathbb R}^n$. Instead, one can as well say "a finite-dimensional  normed real vector space". In fact, a norm is needed here only to define the standard topology on such a vector space, so one could say instead "Hausdorff topological vector space". Which one to use is a matter of taste, as far as I known, there is no difference. The same applies in the case of differentiable manifolds. The only advantage of referring to ${\mathbb R}^n$ is that it simplifies few proofs (for instance, the fact that the tangent space defined via derivations is $n$-dimensional, proof of Morse Lemma, etc.) My attitude to this is to avoid explicit coordinates as much as possible, but use coordinates extensively when working out explicit examples, e.g. proving that a particular subset is a smooth submanifold. Without coordinates, one cannot even ask this question. Another occasion is when proving that, say, $GL(V)$ is a Lie group, where $V$ is a finite-dimensional vector space. Without explicit formulate for matrix multiplication and inversion, this becomes a nightmarish problem. 


One can also approach (potentially) infinite-dimensional differentiable manifolds. For instance, take a look on book by Serge Lang on differentiable manifolds and differential geometry. At the beginning he works without restriction to the finite dimensional case. Then one has to fix a particular topological vector space as a local model. Which class of vector spaces to take depends on the author. Several common choices are Hilbert spaces, Banach spaces, Frechet spaces. They all appear in some interesting examples. For instance, suppose you are studying the space of smooth curves in the Euclidean $n$-space. Depending on your taste, you can consider curves of different degree of smoothness, so you will use different infinite-dimensional model spaces. Another interesting example is the infinite-dimensional Lie group of diffeomorphisms of the circle. or the space of all Riemannian metric on the given finite-dimensional manifold. In each case, you have to do some infinite-dimensional nonlinear analysis, so specifying the model space becomes critical. If you try to say "some topological vector space", you will be unable to prove a single lemma. (To answer your question in comments, Banach spaces are a good choice because you can rely upon, say, the contraction principle. But you may have to be more specific and say, for instance, $W^{2,1}_{loc}({\mathbb R}^2)$. It all depends on the problem you are dealing with.) 


*Geometry might mean "differential geometry". This again can come in different flavors: Riemannian, semi-Riemannian, symplectic, etc. Each of them has its own notion of a "flat space". For instance, in the case of semi-Riemannian geometry of signature $(n,1)$, the flat space can be alternatively described as an $n+1$-dimensional real vector space equipped with a nondegenerate bilinear form of signature $(n,1)$. Or, you can say $R^{n,1}$ with the form
$$
<x,y>= x_1y_1 + ... + x_ny_n - x_{n+1}y_{n+1}.
$$
It is again a matter of personal preference, there is not much difference, until, again you have to deal with examples. There are few things you can prove which are independent of the signature (so you can get away with being "abstract"), but very soon you realize that the theory behaves quite differently depending on the signature: Your elliptic PDEs might suddenly become hyperbolic, 
the semi-Riemannian metric suddenly does not define Hausdorff topology, etc. 

*Or, maybe you want to consider Finsler geometry. Then your flat model space is a finite dimensional normed vector space. Very quickly you again realize the difference between Riemannian and true Finsler setting (e.g. when dealing with geodesics: are they locally unique or not?), curvature, etc.  

*Or, you decide to study symplectic geometery. Then your flat model space is ${\mathbb R}^{2n}$ with a symplectic bilinear form. Instead of saying ${\mathbb R}^{2n}$ you can say real vector space of even dimension, but there is nothing to gain by doing this. 

*Or, geometry might mean "metric geometry", then there is no particular flat model space, but you may have to use say, hyperbolic planes with two different negative curvatures as your "flat model spaces" (in a proof of a single theorem!). 
Lastly, I am not Gowers and I do not know what did he exactly mean, but, likely, what he meant that $S^2$ with the standard Riemannian metric can be described as a "simply connected compact Riemannian surface of constant curvature $+1$". Or, maybe he was thinking about $S^2$ as a topological space, then he could have meant "a compact simply-connected surface (without boundary)". It all depends on the context. 
