Note that $N$ is homeomorphic to the union of $M$ with $\DeclareMathOperator{\Cyl}{Cyl}$ the mapping cylinder $\Cyl(\pi)$ of the bundle projection $\newcommand{\pa}{\partial}$ $\pi:\pa M\to X$. Denote by $M^\circ$ the interior of $M$ Now observe.
Observe that from the Poincare Duality for $M^\circ$ (or equivalently for $(M,\pa M)$ we have) implies
$$
H^{n-k}_{dR}(M)\cong H^k_{dR}(M^\circ) \cong H^k_{cpt}(M^\circ).
$$
The extension by $0$ defines a morphism
$$
H^k_{cpt}(M^\circ)\to H^k_{cpt}(N)\cong H^k(N).
$$$$
H^k_{cpt}(M^\circ)\to H^k_{cpt}(N)\cong H^k(N)\cong \mathrm{Hom}\big(H_k(N),\mathbb{R}\big).
$$
Comment. Above I assumed that $H^\bullet_{dR}(M,\pi)=H^\bullet_{dR}(M)$. Now observe that
$$
H^{n-k}_{dR}(M)\cong H_k(M,\pa M).
$$
From the excision property of homology we deduce that the inclusion
$$
(M,\pa M)\hookrightarrow (N,\Cyl \pi)
$$
induces an isomorphism
$$
H_k(M,\pa M)\cong H_k(N,\Cyl \pi).
$$
If $H_k(X)=H_{k-1}(X)=0$, then $H_k(N)\cong H_k(N,\Cyl \pi)$.