I think, I understand the cause of the confusion. Indeed, every smooth $n$-manifold $M$ is locally diffeomorphic to $R^n$. Hence, every point in $M$ has a (small) neighborhood which admits a flat, i.e., a metric which is isometric to the Euclidean Riemannian metric on (a domain in) $R^n$. Such metrics can be locally characterized y the property that they have zero sectional curvature. (This was Riemann's main discovery.) However, such flat metrics on open subsets of $M$ cannot be (in general) patched together to form a single (flat) Riemannian metric on $M$. The simplest obstruction for this comes from the Gauss-Bonnet formula: If a connected compact surface (without boundary) admits a metric of zero curvature, then this surface has zero Euler characteristic, i.e., is diffeomorphic to the torus or Klein bottle. In particular, the (topological) 2d sphere does not admit a flat metric. Similar things happen in higher dimensions (although the main obstruction is not the Euler characteristic). Much of modern Riemannian geometry revolves around the question of relation between curvature(s) (scalar, Ricci, sectional) and topology of the manifold, i.e., what restrictions on topology of $M$ are imposed by curvature properties of the metric.
Even if one is interested only in local aspects, i.e., Riemannian metrics on open subsets of $R^n$, considering non-flat metrics is still very useful; much of this motivation (going back to Einstein's work) comes from physics.
For your question about the sphere: There is nothing you can do to $R^n$ to make it diffeomorphic to the sphere. Still, you can ask if $R^n$ admits a (complete) metric of strictly positive sectional (or Ricci) curvature (as the round sphere has). The answer to this is negative, but the proof is not so easy, it comes from Myer's theorem: If a manifold $M$ admits a complete metric of strictly positive Ricci curvature, then $M$ is compact. (This is a nice example of interaction of curvature and topology!) Now, observe that $R^n$ is not compact.
Edit: By strictly positive Ricci curvature here I meant that there exists $k$ such that for all $x\in M$, $Ric(x)\ge k>0$. Otherwise, the conclusion is false.