Regarding R-orientability of manifolds

We define the R-orientation for a manifold to be the consistent choice of generators for the R-module $H_n(M|x;R)$ where by consistent we mean each $x\in M$ has a neighbourhood $R^n\subset M$ containing open ball B of finite radius containing x such that the local orientations $\mu_y$ at points $y\in B$ are images of one generator $\mu_B$ of $H_n(M|B;R)\approx H_n(R^n|B)$ under the natural maps $H_n(M|B;R)\rightarrow H_n(M|y;R)$.

Now Hatcher states that any orientable manifold is R orientable for any ring R. A nonorientable manifold is R-orientable iff R contains a unit of order 2.

Now my guess is that the canonical isomorphism $H_n(M|x;R)\approx H_n(M|x)\otimes R$ gives the first statement since given an orientation $\mu_x$ we induce an R-orientation by choosing a generator for R say u and considering $\mu_x\otimes u$. However I am not too sure about this since the canonical isomorphism is a group isomorphism and hence there's no guarantee that the map I defined actually gives a generator for the module $H_n(M|x;R)$.

I would also like to know how the second assertion regarding non orientable manifolds holds.

Your intuition should be right. This canonical isomorphism $H_n(M\mid x;R) \cong H_n(M \mid x; \mathbb{Z}) \otimes R$ does, in fact, provide you with generators: the natural map $\mathbb{Z} \to R$ sends the generators of $\mathbb{Z}$ to generators of $R$ as a rank-$1$ $R$-module.

As for the second assertion, try to think of some non-orientable spaces, and realise that the failure of finding a compatible system of generators always lies in sign issues. Setting $1 = -1$ eliminates precisely that problem.