Intuitive view on $H_2(K)$ where $K$ is the M$\ddot{o}$bius strip Take a Torus $T^2$ and compute its homology group $H_2(T^2)$. It can be deduced that $H_2(T^2) \cong \mathbb{Z}$, and this is correspondent to the intuitive view that the only 2-simplex that is a cycle (boundaryless) and is not a boundary of 3-simplex is the surface itself.

Simiarly, consider a Möbius strip $K$. It can be computed that $H_2(K)\cong\{0\}$. However, in the intuitive view, the surface of $K$ itself is boundaryless, but not a boundary of a 3-simplex, and hence $H_2(K)\cong \mathbb{Z}$.

Is this correct? If it is wrong, which assumption is wrong or which assumption did I miss?
For shure it is something coming from orientability. Applying results of De Rham cohomology you can procide as follows. Retract by a deformation retraction the general Möbius strip to a Möbius strip with fibre $[0,1]$, in this case cohomology with compact support coincides with De Rham cohomology by compactness. Now an application of non-orientability implies that $H_{c}^n(K)$, the cohomology with compact support of $K$, has dimension $0$.  These cohomologies coincides with the singular one in these cases.

Perhaps these arguments are not so satisfying because you need to deal with manifolds and homology theories on manifolds to get a meaning of your computation. However, you can look at the first chapter of "Elements of Algebraic Topology" by Munkres in which you can find a computational approach with triangulations and then prove that the generators of the simplicial homology vanish. As above, in compact topological spaces, singular and simplicial homology are the same.
For shure it is something coming from orientability. Applying results of De Rham cohomology you can procide as follows. Retract by a deformation retraction the general Möbius strip to a Möbius strip with fibre $[0,1]$, in this case cohomology with compact support coincides with De Rham cohomology by compactness. Now an application of non-orientability implies that $H_{c}^n(K)$, the cohomology with compact support of $K$, has dimension $0$.  These cohomologies coincides with the singular one in these cases.

Perhaps these arguments are not so satisfying because you need to deal with manifolds and homology theories on manifolds to get a meaning of your computation. However, you can look at the first chapter of "Elements of Algebraic Topology" by Munkres in which you can find a computational approach with triangulations and then prove that the generators of the simplicial homology vanish. As above, in compact topological spaces, singular and simplicial homology are the same.