$H_0(X) = \mathbb{Z}$ since it's connected.
$H_2(X) = \mathbb{Z}^2$ since there are two non-contractable spheres
$H_1(X) = \mathbb{Z}$ since the torus has two non-contractible curves, but now one can be contracted along the sphere.
Let $a$ be the torus, and $b\cup b' = S^2$ be the upper and lower hemispheres.
$\partial a = a_1 + a_2 - a_1 - a_2 = 0$
Let $a_1 = \mathbb{T}^2 \cap S^2$ be the meridian of the sphere (as well as the torus)
$\partial b = a_1 =\partial b'$
Let $a_{12}$ be a point of $a_1 =\mathrm{torus}\cap \mathrm{sphere}$:
$\partial a_1 = \partial a_2 = \partial b_1 =a_{12} - a_{12}=0$
$H_0 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_{12}\rangle = \mathbb{Z}$
$H_1 = \ker\partial / \mathrm{im} \,\partial = \mathrm{span}\langle a_1, a_2\rangle /
\mathrm{span}\langle a_1\rangle = \mathbb{Z}$
$H_2 = \ker\partial = \mathrm{span}\langle a,b-b'\rangle = \mathbb{Z}^2$
You may also observe the meridian circle $\mathrm{sphere} \cap \mathrm{torus}$ is contractible to a point.
In Hatcher Chapter 0, example 0.8 (page 12) shows us the sphere with two points identified is the homotopy equivalent to the wedge of a sphere and a circle.
$$ S^2 \simeq S^2 \vee S^1$$
For your example, $ \mathbb{T}^2 \cup_{S^1} S^2 \simeq S^2 \vee S^2 \vee S^1$.
Their homology groups will also be the same. $H_0 = \mathbb{Z}, H_1 = \mathbb{Z}, H_2 = \mathbb{Z}^2$.
Let $X$ be a CW-complex whose cellular chain complex $C_{\bullet}(X)$ has all zero differentials. Note that this necessarily occurs when $X$ does not have cells in consecutive dimensions, e.g. $S^n$ for $n \geq 2$, $\mathbb{C} \mathbb{P}^n$ for $n \geq 1$. It also occurs for $S^1$. Then $C_n(X) \cong H_n(X,\mathbb{Z})$ for all $n$. Such a complex must be "minimal" in your sense: if you had fewer cells in any given dimension that would give rise to a smaller Betti number.
It follows from the Eilenberg-Zilber Theorem that a finite product of CW-complexes with all zero differentials has all zero differentials. This shows for instance that the $n$-dimensional torus has this property for all $n \in \mathbb{Z}^+$. In particular this explains all of your examples.
Conversely, let $X$ be a finite CW-complex whose $n$th Betti number is equal to the number of $n$-cells for all $n$. Let $d_n$ be the $n$th differential in the singular complex.
Let $K_n$ be the kernel of $d_n$ and $I_{n}$ be the image of $d_{n+1}$. Then
$K_n \subset \mathbb{Z}^n$, so $K_n \cong \mathbb{Z}^d$ for some $d \leq n$. Further $K_n/I_n \cong \mathbb{Z}^n \oplus T$ for a finite abelian group $T$, so $\mathbb{Z}^n \oplus T$ is a homomorphic image of $\mathbb{Z}^d$. We must then have $d =n$ and $T = 0$, so the map $K_n \rightarrow K_n/I_n$ is a surjective map from $\mathbb{Z}^n$ to itself. Such a thing is known to be an isomorphism, e.g. by structure theory of finitely generated modules over a PID. In other words $I_n = 0$ and all the differentials are trivial. In particular all of the homology groups are free abelian.
Best Answer
As mentioned above, there is no standard notion of "orientability" for CW-complexes. I will assume you meant manifolds instead.
The open Möbius band is a nonorientable manifold. It has the homotopy type of a circle, hence its homology groups are free abelian.
Let $M$ be a compact, non-orientable $n$-manifold. By Poincaré Duality, $\mathbb{Z}/2\mathbb{Z}$-Poincaré Duality and the Universal Coefficient Theorem, both $H_1(M,\mathbb{Z})$ and $H_{n-1}(M,\mathbb{Z})$ have nontrivial $2$-torsion.