Is $\mathbb{R}^2$ with a non euclidian metric a Riemann surface?

Provided that a metric g is regular enough (e.g Chern's condition), one can add a conformal structure to $\mathbb{R}^2$ with metric g to make it a Riemann surface (see also this question).

But I don't understand in the first place why can't we directly state that $\mathbb{R}^2$ is a Riemann surface since the identity atlas with one chart $\mathbb{C} \to \mathbb{R}^2$ is holomorphic (since it has just one chart), independently of g?

Why do we care so much about the metric g (we need to stitch together many isothermal local charts instead), since it doesn't change the topology? In the definition of a Riemann surface we need changes of charts to be holomorphic, and (topological) homeomorphism between $\mathbb{C} \to \mathbb{R}^2$. (Im seeing $\mathbb{R}^2$ as a manifold only with no vector structure/a vector structure induced by its charts). What is mixing me up? Thanks!