Geometric reason as to why $H^2$ of the Klein bottle is $\mathbb{Z}/2\mathbb{Z}$? I was reading this document when I came across the following:

Recall that $H^2(K; \mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.

Here $K$ denotes the Klein bottle. Is there a good geometric intuition/reason as to why this is true? 
 A: The "reason" $H^2(K)\cong\mathbb{Z}_2$ is because the torsion subgroup of $H_1(K)$ is isomorphic to $\mathbb{Z}_2$, which can be explained by the fundamental group. The fact that it shows up in $H^2$ instead of $H^1$ is just an algebraic artifact, a consequence of the universal coefficient theorem. You'd be better off looking at the fundamental group.
A: One of the best ways of understanding integral homology geometrically is through Alexander-Pontryagin duality.  Here is one statement of this duality (see Proposition 3.46 in Hatcher's Algebraic Topology):

Let $M$ be an orientable $n$-manifold, and let $K\subset M$ be compact and locally contractible.  Then for all $i$,
  $$
H^i(K) \,\cong\, H_{n-i}(M,M-K),
$$
  where the coefficients are integral.

In the case of the Klein bottle $K$, the easiest manifold $M$ to use is probably $S^2\times S^1$.  The Klein bottle embeds into $M$ in such a way that its intersection with each meridional $S^2$ is a circle $C$, which rotates $180^\circ$ as one travels around the longitudinal direction.  According to Alexander-Pontryagin duality,
$$
H^2(K) \,\cong\, H_1(M,M-K),
$$
Now, within each meridional $S^2$, the complement of the circle $C$ deformation retracts onto a pair of points.  These points switch as one travels around the longitudinal direction, so $M-K$ deformation retracts onto a single circle that travels twice around the longitudinal direction.  It follows easily that $H_1(M,M-K) \cong \mathbb{Z}_2$.

Edit: Incidentally, it's generally true that $H^n(X) \cong \mathbb{Z}_2$ for a compact non-orientable $n$-manifold $X$.  This can also be proven geometrically using Alexander-Pontryagin duality, where the $M$ is the bundle of $n$-forms on $X$.  This is a line bundle over $X$, with the property that the line “flips” as you go once around any orientation-preserving loop in $X$.  By Alexander-Pontryagin duality,
$$
H^n(X) \cong H_1(M,M-X).
$$
It is easy to show using the long exact sequence for $(M,M-X)$ that $H_1(M,M-X)\cong \mathbb{Z}_2$.  In particular, $M$ deformation retracts on $X$ and $M-X$ deformation retracts onto a 2-fold cover $OX$ of $X$ (the orientation bundle);  the homomorphism $H_1(M-X) \to H_1(X)$ is induced by the covering map $OX \to X$, so its image has index two.
A: Another geometric reason why $H^2(K,\mathbf Z)=\mathbf Z/2$ is the following.
Recall that there is a universal integral second cohomology class $u\in H^2(K(\mathbf Z,2),\mathbf Z)$, where $K(\mathbf Z,2)$ is the Eilenberg-MacLane space. Note that $K(\mathbf Z,2)$ is nothing but the inifinite complex projective space $\mathbf P^\infty(\mathbf C)$. Universality means then that the pull-back morphism
$$
\begin{align*}
[K,\mathbf P^\infty(\mathbf C)]&\rightarrow H^2(K,\mathbf Z)\\
f&\mapsto f^\star u
\end{align*}
$$
is a bijection, where $[\cdot,\cdot]$ denotes the set of homotopy classes of maps from the first $\cdot$ into the second. Since $K$ is a $2$-dimensional CW-complex, the $2$-skeleton of $\mathbf P^\infty(\mathbf C)$ is the sphere $S^2$, and $\mathbf P^\infty(\mathbf C)$ has no $3$-cells, one has
$$
[K,S^2]=[K,\mathbf P^\infty(\mathbf C)].
$$
To put it otherwise, the elements of $H^2(K,\mathbf Z)$ correspond to homotopy classes of maps from $K$ into $S^2$.
Now, the Klein bottle is a union of two Moebius strips glued together along their common boundary. Let $E$ and $E'$ be the equators of the two Moebius strips. The surface obtained from $K$ by contracting $E$ to a point, and by contracting $E'$ to another point is the sphere $S^2$. This gives a continuous map
$$
f\colon K\rightarrow S^2
$$
whose topological-degree-mod-$2$ is nonzero. It gives rise, therefore, to a nonzero element $f^\star u$ of $H^2(K,\mathbf Z)$. This element is, moreover, a generator of $H^2(K,\mathbf Z)$. In order to see that $2f^\star u$ is equal to $0$, suppose that $f(E)$ is the North pole, and $f(E')$ is the South pole of $S^2$. Let $s\colon S^2\rightarrow S^2$ be the square map $s(x+iy,z)=((x+iy)^2,z)$, considering $S^2$ as a subset of $\mathbf C\times\mathbf R$. Then it is easy to see that $s\circ f$ is null-homotopic. Then $2f^\star u=(s\circ f)^\star u=0$ in $H^2(K,\mathbf Z)$, and $H^2(K,\mathbf Z)=\mathbf Z/2$.
In a nutshell, $H^2(K,\mathbf Z)=\mathbf Z/2$ since there are only two homotopy classes from the Klein bottle $K$ into the sphere $S^2$.
A similar argument applies to any nonorientable compact connected topological surface without boundary.
