If you know that a shape tiles the plane, does it also tile other surfaces?

You are asking several questions, I understand only the first one, the rest will require some major clarification before they become answerable:

Question 1. Let $M$ is a Riemannian surface homeomorphic to the plane. Does $M$ admit a tiling? Here a tiling means a partition of $M$ into pairwise isometric relatively compact regions with piecewise-smooth boundary, such that two distinct tiles intersect along at most one boundary curve.

This question has a very easy an negative answer. For instance, start with the Euclidean plane $E^2$ and modify its flat metric on an open ball $B$, so that the new metric has nonzero (at some point) curvature in $B$ and remains flat (i.e., of zero curvature) outside of $B$. (This modification can be even made so that the surface $M$ is isometrically embedded in the Euclidean 3-space $E^3$: start with the flat plane in $E^3$ and make a little bump on it.) The resulting manifold admits no tiling, since all but finitely many tiles would be disjoint from $B$ and, hence, have zero curvature metric. At the same time, at least one tile $T_1$ will contain points where the curvature is not zero. Therefore, $T_1$ cannot be isometric to the tiles which are disjoint from $B$.

The same (or similar) construction can be used for any (connected, noncompact) smooth 2-dimensional manifold $S$, no matter what its topology is. This manifold always admits a metric $g_0$ of constant curvature $-1$. Now, modify the metric $g_0$ on a small ball $B\subset S$ making it into a metric of nonconstant curvature in $B$. Since $S$ is non-compact, the same argument as above shows that the new Riemannian manifold admits no tiling. If $S$ is compact, then, of course, each metric on $S$ admits a tiling, consisting of a single tile. If you require at least two tiles, one can always construct metrics which do not admit tilings, but this, requires more work.

However, if your manifold $M$ is, say, a flat 2-torus, i.e., genus 1 closed surface equipped with a zero curvature metric, then such $M$ always admits a tiling consisting of more than one tile. Let me know if you want to see a proof (it is quite easy).

I believe you mean 2-sphere, which is a 2-d surface that lives in 3-d space. None of the tilings of the plane extend to the 2-sphere because of the Euler characteristic. You can tile the 2-sphere with pentagons, but not with squares or hexagons (except for the trivial case of dividing a great circle into 4 or 6 segments and using two squares or hexagons.)