The short answer is: You can't give $\mathbb{S}^2$ a flat metric, because the metric must be assigned to each point in a smooth manner. The different ways in which this can be done are subject to topological constraints. The most famous and elementary example of this is the Gauss-Bonnet theorem, which says that for any closed surface $S$ and any smooth Riemannian metric $g$ on $S$, its sectional curvature $\kappa_g$ must satisfy
$$\int_S \kappa_g = 2\pi\chi(S).$$
(Here, $\chi(S)$ denotes the Euler characteristic of $S$.)
An important class of problems in Riemannian geometry is to understand the interaction between the curvature and topology.
An example of such interaction is given by the Gauss-Bonnet theorem which relates the geodesic curvature, the Gaussian curvature to the Euler characteristic of a regular surface of class $C^3$.
When studying the geometry of a smooth manifold we need to introduce the commutator of twice covariant differentiating vector fields which is called Riemannian curvature tensor.
As a matter of fact, if in Euclidean space we can change the order of differentiation, on a Riemannian manifold, the Riemann curvature tensor is in general nonzero.
For surfaces, the Riemann curvature tensor is equivalent to the Gaussian curvature K, which is scalar function.
On the other hand in dimensions larger than two, the Riemann curvature tensor is a tensor–field.
Now there are several curvatures associated to the Riemann curvature tensor.
Given a point $p \in M^n$ and two dimensional plane $\Pi$ in the tangent space of M at p, we can define a surface S in M to be the union of all geodesics passing through p and tangent to $\Pi$.
In a neighborhood of p, S is a smooth 2D submanifold of M. Then it is possible to define the sectional curvature $K(\Pi)$ of the 2D plane to be the Gaussian curvature of S at p.
Thus the sectional curvature $K$ of a Riemannian manifold associates to each 2D plane is a tangent space a real number.
You can imagine the sectional curvature as a kind of generalization of the Gaussian curvature.
Best Answer
It's the metric. Note that it need not make sense to apply a Euclidean inner product to tangent vectors. If you see $\langle\cdot,\cdot\rangle$ in the context of a Riemannian manifold, it's the metric.