Your definition of Euler characteristic does not agree with the notes you link to. Nicolaecu's notes talk about compactly supported cohomology groups, and it is true (as he calculates, that $H_c^k(R^n) \cong 0$ if $k\neq n$ and $\cong R$ if $k=n$, hence $\chi(R^n)=(-1)^n$ with his definition of Euler characteristic (there is also another notion (and more commonly used) of Euler characteristic, which is the one you define, but it is different from the compactly supported one)
I don't know how intuitive this will be, but here is how I think of it. It is similar to what Travis mentioned in the comments. We need the Euler characteristic to not change when we change the triangulation. Consider the following triangulation of the plane.
This graph has 4 vertices, 5 edges, and 3 faces, which gives us the $4-5+3=2$ we expected. Now, if we were to pretend that we don't know how to compute the Euler characteristic, and just that we don't want it to change if we change the triangulation, we see that the new edge and the new vertex need to cancel each other out.
We also need this to work for the faces, so if we remove an edge, we will decrease the number of edges and decrease the number of faces. We still don't want it to change.
So, edges and faces also need to cancel each other out. The easiest way to get the Euler characteristic to remain invariant under these changes is to have the even dimensional count positively and the odd dimensions be negative. Like I said, this is how I think of it, hopefully it helps your intuition too.
Best Answer
The Euler's characteristic is a topological invariant, namely two homeomorphic spaces have the same Euler's characteristic but is not a total topological invariant. It does not suffice that two topological spaces have the same Euler's characteristic to ensure they are homeomorphic.