To be honest, I am not really sure what is your question. What do you mean by "every graph has a genus"? I'll try to answer by the way, since geometrically your question is clear.
For me, the genus is a well defined quantity associated to a (compact) surface. When you have a planar graph, you imagine to fill the faces among the edges and you get a very simple surface, something like a polygon in the plane. If you also fill the exterior " ghost" face you obtain the whole plane. The genus is computed now as $1-\chi/2$, where
$$\chi = \#\{\text{filled faces}\} - \#\{\text{edges}\} +\#\{\text{vertices}\}$$
And we would be done, but the plane is not a compact surface so we don't know what this should be equal to. To deal with this technical issue, we add a point to infinity and we get a sphere. Now the theory tells us that the genus (which is secretly the number of holes) of the sphere is zero, so that we get for a planar graph
$$\#\{\text{filled faces}\} - \#\{\text{edges}\} +\{\text{vertices}\}= 2$$
You can also reformulate this, recalling that the faces we filled were the internal faces + one ghost face:
$$\#\{\text{ faces}\} = \#\{\text{edges}\} - \{\text{vertices}\}+1$$
Check this on small graphs! That's true for real :)
Some complications arise if there are intersections. The main idea in algebraic topology is that if we cut a topological space in simple bricks - like a surface in triangles - the geometric properties of the space become combinatorial properties of how our bricks are combined. You have just seen this above: the genus of the sphere reflects in an equation for edges, vertices and faces for the graph that "builds up" the sphere.
When you have intersection, the problem is that the graph is apparently not building up anything. The idea of your book is to create small bridges (handles) so that the graph does not intersect himself anymore: you will have a planar graph sitting inside a new surface, no longer a sphere, but who cares! We can carry on our argument above, and we only have to change the genus in the end: for a nice coincidence, it will be the number of bridges we added.
If you are not comfortable with deforming the surface into a k-hole torus while preserving planarity, there is no need to do that: the genus is a homotopy invariant. If you are convinced that - beside the graph story - you can deform a k-handle sphere into a k-hole torus, then you know that the genus of the k handle sphere is the same of the k-hole torus, that is k.
Please ask in the comments if you need further clarifications :)
Best Answer
First of all, the terminology "2-torus" is incorrect. The term $n$-torus refers to a torus of dimension $n$. What you mean, I think, is a genus-2 closed orientable surface.
Now, consider a genus-2 surface that looks like O==O (the two holes on either side, with a tube connecting them). Cut it through the middle like so, O=|=O --> O= + =O. Now you have two punctured tori, or, as you explained above, two tori each missing exactly one 2-handle, or 0-handle, depending on which side you're looking from). Since you want to look from bottom to top, let's say the first one has decomposition 0-handle + 1-handle + 1-handle, and the second one has 1-handle + 1-handle + 2-handle. Put them back together and there's your decomposition, without having to worry about 4-punctured spheres.
However, just for fun, let's try to come up with the handle decomposition of a 4-punctured sphere (by the way, this is sometimes called a "pair of monkey pants" -- the waist, two holes for the legs, one for the tail). A disk is a once-punctured sphere, so let's start with a 0-handle. Adding a single one-handle gives us, topologically, an annulus, or a two-punctured sphere. If you see the pattern, let's add two more one-handles to up the puncture number to 4, and we're done. A 0-handle + three 1-handles is a 4-punctured sphere.