A form $\beta^{p}$ is closed if $d\beta=0$.
A form $\beta^{p}$ is exact if $\beta^{p}=d\alpha^{p-1}$, for some form $\alpha^{p-1}$.
An orientable closed manifold is a compact manifold without boundary.
What is a compact manifold?
Why is an orientable closed manifold the same as a compact manifold without boundary?
If a manifold is compact, does it not, be definition, have a boundary?
How do you prove that the integral of an exact form over an orientable closed manifold is $0$?
I will try to answer your questions:
1) A manifold is compact if every open covering (i.e. a collection of open sets that contains the manifold in its union) has a finite subcovering. So from this open covering, which might have infinitely many sets, you can choose finitely many and these must still cover the manifold.
2)On this website https://en.wikipedia.org/wiki/Classification_of_manifolds a closed manifold is defined as a compact manifold without boundary.
3) No, a compact manifold could have no boundary. For example, consider the circle $S^1$. It is definitely compact and does not have a boundary.
4) As we have just concluded that this manifold has no boundary, we can just use Stoke's theorem for our manifold $M$: \begin{equation} \int_{M} d\omega = \int_{\partial M} \omega. \end{equation} We have an exact form, let's call it $\beta^p$. As you already said, this means that $\beta^p=d\alpha^{p-1}$. Also, because $M$ has no boundary, the integral over $\partial M$ will be zero: \begin{equation} \int_M \beta^p = \int_M d\alpha^{p-1} = \int_{\partial M} \alpha^{p-1} =0. \end{equation}