Let $M$ be a compact orientable $n$ dimensional smooth manifold without boundary, and $e\in H^n_{dR}(M)$ denote the Euler class of the tangent bundle $TM$. We have that \begin{align*} \int_Me=\chi(M) \end{align*} where $\chi(M)$ is the Euler characteristic.
Now, I know of a couple constructions of the Euler class, both of which are mildly involved. One involves the sphere bundle induced by a vector bundle, and the other involves pulling the Thom class back by a section. Both of these makes sense to me, but what I am confused about is whether in the case of the Euler class of the tangent bundle, if it is correct to state that the Euler class is $\chi(M)$ times a generator of $H^n_{dR}(M)$ (by generator, I mean a top form which integrates to $1$).
I feel like this is incorrect since I think that many forms should be able to integrate to $\chi(M)$, and we do so much to construct the Euler class. Then again in the case of a compact orientable smooth manifold we have that integration is an isomorphism from $H^n_{dR}(M)\cong \mathbb{R}$, so I guess any form that happens to be closed and integrates to zero is also actually exact, which is mildly shocking when I think about it.
So, is it correct to say that the Euler class is $\chi(M)$ times a generator of $H^n_{dR}(M)$?
For context, I am reading a paper where the author looks at the Euler class of the vertical tangent bundle of a fibre bundle (with compact orientable boundaryless fibres), and then states that the restriction of the Euler class to the fibre is the Euler characteristic times a generator of the top cohomology of the fibre. Shouldn't this just be the Euler class? Especially since the fibres are embedded submanifolds, so the restriction is just pulling back by the inclusion map, and the functorality of the Euler class should guarantee that this form pulls back to Euler class.
Am I missing something?