There is a huge interaction between the Riemannian metric of a manifold and its topology. This interaction appears through its curvature.
A first example is the well-known Gauss-Bonnet formula. Given a Riemannian surface $(\Sigma,g)$, denote by $K$ its Gaussian curvature. Then:
$$\int_\Sigma K = 2\pi\chi(\Sigma)$$
Thus, the metric tensor is forced to have a global behaviour controlled by the topology of the surface. As an application, there does not exist any metric with negative curvature on the sphere or on the torus.
Another well-known interaction is the following. If $(M,g)$ is a Riemannian manifold with non-positive sectionnal curvature, then its universal cover is diffeomorphic to $\mathbb{R}^n$. This implies that a $n$ dimensional sphere does not have any metric of non-positive sectionnal curvature.
A last interaction is more subtle: the bahaviour of the curvature affects the aglebraic properties of the fundamental group of a compact manifold. On some positive assumptions on the curvature, the fundamental group has to have polynomial growth, while with some negative assumptions, it has at least exponential growth.
Note that every manifold can be endowed a Riemannian metric. The case of pseudo-Riemannian metrics is not the same because one cannot glue local pseudo-Riemannian metrics with partition of unity and expect the result to be a pseudo-Riemannian metric. For example, this theorem shows that the Lorentzian case is restricted to a few examples: every noncompact connected smooth manifold admits a Lorentz metric, and a compact connected smooth manifold admits a Lorentz metric if and only if its Euler characteristic is zero.
In your example, the manifold $S^2\times\mathbb{R}^2$ is simply connected and is not diffeomorphic to $\mathbb{R}^4$. Hence, it does not admit a metric with negative curvature.