The book Introduction to Topology by C. Adams and R. Franzosa says :
From the triangulations in Figure 14.8, we see that $\chi(S^2) = 2$, $\chi(T^2) = 0$, $\chi(K) = 0$ and $\chi(P) = 1$.
And here is the "Figure 14.8" :
I tried about 2 hours for only $S^2$ (!!) and every time I got different numbers esp. non of them were $2$. I used different tricks with a very high caution for what happens for vertices and edges of 'boundary-triangles' but I failed.
I know an alternative but very easy way to calculate $\chi(S^2)$ which is cutting the $S^2$ (with origin as the center) by surfaces of $x=0$, $y=0$ and $z=0$ so $\chi(S^2)=F+V-E=8+6-12=2$.
And about calculating $\chi(T^2)$, I cut the $T^2$ (with origin as the center) by surfaces of $z=0$ and $y=0$ then having $4$ 'rectangles' and cut each of them by their diameters to two pieces so $\chi(T^2)=F+V-E=8+4-12=0$.
Being deprived of the ability to visualize a $4$ dimensional space, I cannot operate the same easy procedure to the embedding of $K$ and $P$ in $\mathbb R^4$.
Truly I would appreciate any help for a simple clear way to calculate $\chi(K)$ and $\chi(P)$. Thank you.
I apologize for my paint abilities, but consider the following colored image corresponding to the description of $K$ as quotient:
The triangulation of $K$ described by the drawing has nine vertices and not sixteen. Remember that $K$ is a quotient of the square by the identification depicted through the bold arrows. For example, the four vertices of the big square (colored in the same color in the picture above) are identified by the quotient map and so count as one vertex for the triangulation of $K$. Similarly, all the colored vertices and edges that have the same color are identified by the quotient map and must be counted accordingly.