Q:

I'm looking for a proof for the Euler Characteristic of non-orientable surfaces, $χ=2−k$

Unfortunately my notes only cover the proof up to orientable surfaces, i.e., $χ=2−2g$, where $g$ is the genus. I would like to get some guidance on where to search for this proof. I have tried looking it up on Topological Graph Theory, by J. L. Gross and T. W. Tucker, but could not find it anywhere.
I have found some questions on this site asking about this, but they make reference to homology theory or something called cellularisation, which I have no idea of.
Edit: Explanation of the terms
Here, $χ= n - m + f$ where $n$ is the number of vertices, $m$ the number of edges, and $f$ the number of faces.
$k$ is the genus of the non-orientable surface.

A:

All you need to do is to construct a single cell decomposition of the nonorientable surface of genus $k$ --- meaning a decomposition into vertices, edges and faces --- count the number $n$ of vertices, $m$ of edges, and $f$ of faces, compute $n-m+f$, and verify that the result is $2-k$. 
Now, there are many, many, many different cell decompositions, and you only need one, because they all give the same answer. If you ever wonder why that's true, that's what you need homology to prove. But, setting that aside, all you need is one decomposition.
The easiest cell decomposition that I know of comes from gluing the sides of a $2k$ sided polygon in the pattern $a_1a_1a_2a_2a_3a_3....a_ka_k$. One checks that all vertices are identified to a single vertex so $n=1$, the $2k$ sides are identified to $k$ edges so $m=k$, and the interior of the polygon gives a single face so $f=1$, hence $n-m+f=2-k$. 
I encourage you to try to find your own cell decomposition, count, and compute.

