But the devil is in the details. Assume each connected region of constant elevation contracts to a point. This does not disturb the topology of the sphere. Note that certain troublesome examples are ruled out, like a perfectly uniform sphere with no hills, valleys, or passes. Similarly, a plateau cannot look like a ring, since that is not simply connected, and will not contract to a point. If the equator bulges out from the rest of the sphere then the sphere has one hill and two valleys (at the poles), hence h-p+v is 3 instead of 2.
Another problem is a pass between more than two hills. Start with a well behaved topography, and pull one of the hills apart into three, with one pass between them. This adds two hills and one pass, and that's going to cause trouble. Turning to the equivalent graph, three vertices are joined by one edge. Thus each critical point is a hill, or a valley, or a pass that looks like a saddle point with one direction curving up and the perpendicular direction curving down. This is assured when the surface is approximated locally by a quadratic function.
Another requirement is finitely many critical points; otherwise we can't apply euler's formula.
Now let's draw the graph. As mentioned earlier, every hill becomes a vertex. Start at a pass and move in either of the two opposite directions that is concave up. Follow the gradiant until you reach the next critical point. The path that follows the gradiant is called an integral curve, and it is unique. (This is a theorem from differential geometry, and it assumes a continuous gradiant, which is used to construct the path.) This explains why edges can't intersect. Each point with a nonzero gradiant establishes the path uniquely, going forward and backward in time. Separate edges cannot intersect, and an edge cannot cross itself.
Climb the integral curve until you reach another critical point, which has to be a hill or a pass. Yes, a pass can lead to another pass. Start at one pass, with a valley to your left and right, and walk straight ahead, climbing up the mountain until you reach another pass, with a hill to your left and right. This becomes an edge that runs smack into another edge, without a connecting vertex. This is not a problem, thanks to a variant of euler's formula. So - draw an edge for each pass, and you have satisfied the conditions for an embedded graph.
Start at a valley an travel in any direction, following the integral curve. You eventually reach a critical point, which is a pass or hill. Thus each vally is a face, surrounded by a ring of edges and vertices. This completes the correspondence to the entities in euler's formula.
For an arbitrary 2 manifold, h-p+v = 2-2g.
If a pass is nestled among many hills, push up on the pass slightly, to make a new hill. This produces a valid graph. Pull the hill back down, and we have lost one vertex and two or more edges. This increases the value of euler's formula. If you don't know that each pass is approximately quadratic, you can still write h-p+v ≥ 2-2g.