Hi. Is there a result in the spirit of Poincaré duality but for nonoriented manifolds? Thanks.

Yes. For simplicity, assume $M^n$ closed and connected. For each $x \in M$, let $\omega_{M,x}:= H_n (M; M \setminus x)$. This is a coefficient system of groups isomorphic to $\mathbb{Z}$ (or any other ring, if you start with that). Now you can observe that $H_n(M,M\setminus x; \omega_M)$ is \emph{canonically} isomorphic to $\mathbb{Z}$ (the isomorphism depends on the choice of an isomorphism $H_n (R^n, R^n \setminus 0 ) \cong Z$. The customary proofs of Poincare duality now show: there exists a unique element $[M] \in H_n (M; \omega_M)$ which restricts to the given generator in $H_n (M, M \setminus 0;\Omega_M)$ for each $x \in M$. cap product with the fundamental class induces a isomorphisms $H^{k}(M; \omega_M) \to H_{nk}(M; \mathbb{Z})$ and $H^{k}(M; \mathbb{Z}) \to H_{nk}(M;\omega_M)$. An orientation is by definition a trivialization of $\omega_M$; from that you recover the oriented Poincare duality theorem, with the familiar dependence on the orientation. 


A high level purely homotopical answer that does not require explicit consideration of homology or cohomology is given in Theorem 19.1.5 of Parametrized Homotopy Theory, by Johann Sigurdsson and myself. That result works equivariantly, where orientation is not as well understood as one would like. Nonequivariantly, the result gives a description of the spectrum $k\wedge M_+$ for any spectrum $k$ and smooth closed manifold $M$ as a function spectrum defined in terms of parametrized spectra. That may sound daunting, but it is really very natural. Here $k$ doesn't even have to be a ring spectrum. 


Another kind of answer is Atiyah duality: if we write $D(X)$ for the SpanierWhitehead dual of $X$ then $D(M/\partial M)$ is the Thom spectrum $M^{TM}$. For any finite spectrum $X$ we have $H_k(DX)=H^{k}(X)$ and $H^k(DX)=H_{k}(X)$. For any finite complex $X$ and any orientable virtual vector bundle $V$ of virtual dimension $d$ we also have $H^k(X^V)=H^{kd}(X)$ and $H_k(X^V)=H_{kd}(X)$. If $M$ is an orientable manifold of dimension $n$ this gives $H^{kn}(M^{TM})=H^k(M)$. By combining these facts we can recover Poincaré duality from Atiyah duality. 

