Rational cohomology of closed nonorientable manifolds with coefficients in the orientation sheaf. Let $M$ be a closed nonorientable manifold. I know the rational cohomology of manifolds over trivial coefficients $\mathbb{Q}.$ I want to know about the rational cohomology of $M$ with coefficients in the orientation sheaf $\mathbb{Q}^{w}$. I am a beginner in algebraic topology. Is there any simple definition of orientation sheaf $\mathbb{Q}^{w}?$ How do I relate $H^{*}(M;\mathbb{Q}^{w})$ with $H^{*}(M;\mathbb{Q})?$ Moreover, is there any simple understandable reference for the orientation sheaf $\mathbb{Q}^{w}?$ 
 A: Since you're a beginner at Algebraic Topology this topic might be a bit difficult (even as a grad student in algebraic topology I always found twisted coefficients to be difficult/hard to work with explicitly). Kirk and Davis' book "Lecture Notes in Algebraic Topology" has a chapter covering homology with local (aka twisted) coefficients starting on page 97. Here is how I managed to understand the relation between $H^*(M;\mathbb{Q}^\omega)$ and $H^*(M;\mathbb{Q})$:
Claim: there is a twisted version of Poincaré Duality with rational coefficients. Kirk and Davis state it for $\mathbb{Z}$ on page 104, and then below state that there is a generalization for any $\mathbb{Z}[\pi_1(M)]$ module $A$, and we can take $A=\mathbb{Q}$ where $\alpha\in\pi_1(M)$ acts by $-1$ if the tangent bundle is non-orientable over a loop representing $\alpha$, and acts trivially otherwise. In particular, if $M$ is a closed, connected manifold of dimension $n$ and $\mathbb{Q}^\omega$ is the coefficient system twisted by the orientation character then there is a twisted fundamental class $[M] \in H_n(M; \mathbb{Q}^\omega) \cong \mathbb{Q}$ inducing isomorphisms
$$ - \cap [M]\colon H^k(M;\mathbb{Q}) \cong H_{n-k}(M;\mathbb{Q}^\omega) $$
$$ - \cap [M]\colon H^k(M;\mathbb{Q}^\omega) \cong H_{n-k}(M;\mathbb{Q})$$
Remarkably, if $M$ is non-orientable this implies that $H^0(M;\mathbb{Q}^\omega) \cong H_0(M;\mathbb{Q}^\omega) \cong 0$ (as Captain Lama points out in their comment, the intuition behind this is that the orientation sheaf has no global section when the manifold is non-orientable).
Now, since $\mathbb{Q}$ is injective as a $\mathbb{Z}$-module, the Universal Coefficient Theorem gives us an isomorphism $\kappa\colon H^k(M;\mathbb{Q})\cong Hom(H_k(M;\mathbb{Z}), \mathbb{Q})$. But $$Hom(H_k(M;\mathbb{Z}), \mathbb{Q}) \cong Hom(H_k(M;\mathbb{Z})\otimes \mathbb{Q}, \mathbb{Q}) \cong H_k(M; \mathbb{Z})\otimes\mathbb{Q}\cong H_k(M;\mathbb{Q})$$
Here the first isomorphism is because a homomorphism to $\mathbb{Q}$ will kill any torsion so we can ignore it by tensoring with $\mathbb{Q}$, the second isomorphism is because they are vector spaces, and the third isomorphism is by the Universal Coefficient Theorem for Homology and uses the fact that $\mathbb{Q}$ is torsion-free.
Combined together we actually find
$$H^k(M;\mathbb{Q}^\omega) \cong H_{n-k}(M;\mathbb{Q})\cong H^{n-k}(M;\mathbb{Q}) $$
