This is a closed orientable 3-manifold so its homology groups are determined by its fundamental group, which is the binary octahedral group. There is a very nice book by Montesinos that covers all examples of this type in three dimensions, called "Classical Tessellations and Three-Manifolds".

The question should be rephrased to eliminate the ambiguity of what the word "its" refers to. (Does it refer to "connected component" or to "covering space"?)

In the same spirit, just take the smash product of two CW complexes that are Moore spaces for finite cyclic groups of relatively prime orders.

The question as originally stated assumes the two manifolds have different dimensions. If this is what was intended, then the manifolds cannot both be closed, and counterexamples are easy to find. For example, let $M$ be the circle and $N$ the Moebius band (either the compact or the noncompact version). These have different classes $w_1$ since $M$ is an orientable manifold and $N$ is not. Higher-dimensional counterexamples are equally easy to find.

The stable parallelizability of exotic spheres is Theorem 3.1 of the famous Kervaire-Milnor paper "Groups of homotopy spheres I" in the 1963 Annals. The proof is short but uses several big theorems from the previous decade such as Bott periodicity, the Hirzebruch signature theorem, and Adams' work on the J-homomorphism.

For $BO(3)$ one can choose the Grassmannian of 3-planes in ${\mathbb R}^\infty$, and this is in some sense approximated by the finite-dimensional Grassmann manifolds of 3-planes in ${\mathbb R}^n$. What kind of approximation do you have in mind?

The third condition in the definition of a polyhedral surface currently reads: "If an edge of a polygon in C intersects an edge of another polygon in C in a common vertex, then the two edges are also edges of a third polygon in C." This seems too restrictive since it excludes vertices of valence greater than three. Lee Mosher's answer gives the correct condition.