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 $\mathbb{Q}^{w}?$
Rational cohomology of closed nonorientable manifolds with coefficients in the orientation sheaf.
algebraic-topologyhomology-cohomologysheaf-theory
Related Solutions
Chern classes can be defined in multiple different but roughly equivalent ways. As you have seen, they can be defined to be certain classes in $H^{2i}(X;\mathbb{Z})$, or they can be defined to be certain classes in $H^{2i}(X;\mathbb{R})$ (in the latter case, usually because they arise naturally via de Rham cohomology). However, these two definitions are very closely related and there is usually no confusion in calling both of them "Chern classes".
To relate the definitions, recall that any space $X$, there is a natural homomorphism $H^*(X;\mathbb{Z})\to H^*(X;\mathbb{R})$, induced by the inclusion map $\mathbb{Z}\to\mathbb{R}$. The two types of Chern classes mentioned above are related by this natural homomorphism: if you take the Chern class in $H^{2i}(X;\mathbb{Z})$ and apply the natural homomorphism $H^{2i}(X;\mathbb{Z})\to H^{2i}(X;\mathbb{R})$, you get the Chern class in $H^{2i}(X;\mathbb{R})$. When people talk about classes in de Rham cohomology as actually being in $H^{*}(X;\mathbb{Z})$, they really mean they are in the image of this homomorphism.
In general, beware that the homomorphism $H^{2i}(X;\mathbb{Z})\to H^{2i}(X;\mathbb{R})$ may not be injective (the kernel is the torsion elements of $H^{2i}(X;\mathbb{Z})$), and so the integral Chern classes are "stronger" than the real Chern classes. For instance, a complex line bundle is determined up to isomorphism by its first integral Chern class, but it is not determined by the first real Chern class (if $H^2(X;\mathbb{Z})$ has torsion).
Yes, $H^{n-1}(M-\{p\};\mathbb{Q}) \neq 0$ for any closed non-orientable manifold of dimension $n$. Long-story-short it's because $H^n(M)=0$ in this case so $H^{n-1}(M-\{p\})$ surjects onto the coefficient group.
Just for exposition I wanted to spell out the arguments to a few of your assertions. Consider the long exact sequence of the pair:
$$\dots \to H^{k}(M, M-\{p\}) \to H^k(M) \to H^k(M-\{ p\})$$$$ \to H^{k+1}(M, M-\{p\}) \to H^{k+1}(M)\to \dots $$
This sequence exists for all coefficients $R$, in particular $\mathbb{Q}$.
Since $M$ is a manifold of dimension $n$, the group $H^k(M, M-\{p\})$ is $R$ if $k = n$ and $0$ otherwise. It follows that $$H^k(M) \cong H^k(M-\{p\})\text{ if $k < n-1$}$$
and there is an exact sequence
$$0 \to H^{n-1}(M) \to H^{n-1}(M-\{ p\}) \to R \to H^n(M) \to 0$$
where $H^n(M-\{p\}) = 0$ because it is an open manifold.
If $M$ is orientable then the last map is an isomorphism so $H^{n-1}(M) \cong H^{n-1}(M-\{p\})$. Otherwise $M$ is non-orientable, so if $R$ is such that $H^n(M) =0$ (for example $R=\mathbb{Z}, \mathbb{Q}$) then you get a short exact sequence with $H^{n-1}(M-\{p\})$ in the middle: if $R=\mathbb{Q}$ that means $H^{n-1}(M-\{p\})$ is a rational vector space of dimension $dim(H^{n-1}(M)) +1>0$. In fact as long as $H^n(M)=0$ and $R\neq 0$ we must have $H^{n-1}(M-\{p\};R)\neq 0$ because it surjects onto $R$.
Best Answer
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
$$H^k(M;\mathbb{Q}^\omega) \cong H_{n-k}(M;\mathbb{Q})\cong H^{n-k}(M;\mathbb{Q}) $$