What made the proof of the four color theorem on planes so hard? What made the proof of the four color theorem for planar graphs so hard? Analogous theorems on different objects (e.g. the torus) were proven long before the planar (spherical) case. Why was the planar case so hard?
Or considering that "why" isn't a very well-defined term in math, what was the obstacle in the proof of the planar case that doesn't come up in non-planar cases?
 A: To give a concise answer, we have a few theorems that provide upper bounds for the coloring number of surfaces. For many surfaces this upper bound makes proving the coloring number of a surface very easy. (Suppose the coloring number of some surface is at most $n$ and we find a map on this surface that requires $n$ colors, then we are done). Unfortunately the upper bound we get from these theorems for the plane (which turns out to be six) isn't as close to the true coloring number (four) that we would hope, which is why proving the coloring number for the plane is so much harder than for other surfaces.

To give a longer answer, let's actually look at the theorems and look at why they don't help us as much in the planar case. 

Given a surface $S$, let $N$ be the smallest integer such that for any complex on $S$ with $e$ edges and $f$ faces, we have $\frac{2e}{f} < N$. Then $N$ colors are sufficient to color any map drawn on $S$.

The idea behind the proof of this statement is that we look at any complex on $S$ and take the set of polygons that form that complex, there will be $f$ polygons with a total of $2e$ edges (because two edges are identified while building the complex). So $\frac{2e}{f}$ is like the average number of edges per polygon. We can do induction on $f$ to show that this average is less than $N$. But the quantity $\frac{2e}{f}$ depends on the complex imposed on $S$. It would be nice to have a theorem bounding the coloring number of a surface independent of the complex.

Given a surface $S$ with Euler characteristic $\chi(S)$, for any complex on $S$ we have
  \begin{equation}\tag{1}
  \frac{2e}{f} \leq 6\left(1 - \frac{\chi(S)}{f}\right)
\end{equation}

Now we can combine these two theorems and start computing the Euler characteristics of some familiar surfaces to find out how many colors would be sufficient to color any map on that surface.
\begin{array}{|c|c|c|}\hline
  \text{Surface ($S$)} & \chi(S) & \text{Sufficient Number of Colors by (1)} \\\hline
  \text{The plane $\mathbb{R}^2$ (or sphere $\mathbb{S^2}$)} & 2 & 6 \\
  \text{The torus ($\mathbb{T}$)} & 0 & 7 \\
  \text{The Klein bottle} & 0 & 7 \\
  \text{The real projective plane ($\mathbb{R}P^2$)} & 1 & 6 \\\hline
\end{array}
While these values just give us an upper bound on the coloring numbers, for some of these surfaces it is also necessary to use that many colors. For the torus and for the real projective plane we can come up with maps that require the number of colors given by (1), meaning that we've found the coloring numbers of these surfaces. For other surfaces the coloring number is actually lower. The true coloring number of the Klein bottle is $6$, so we would have to refine our upper bound a bit. The true coloring number of the plane is $4$, so refining the upper bound is a bit more arduous. Proving that $5$ is actually an upper bound on the coloring number of the plane isn't too bad. It's refining the upper bound from $5$ to $4$ that turns out to be the difficult problem, and is the reason that the four color theorem so (in)famous. You can read a sketch of these refinements in David Speyer's answer here.
For more details on this idea see Kinsey's Topology of Surfaces, Chapter 5. Also read about the Heawood number of a surface.
