When can a manifold written as boundary of a higher dim manifold? From Stoke's we have, we have:
$$\int_{\partial \Omega} w = \int_{\Omega} dw$$
Does it make sense to write, $\Omega= \partial \phi$:
$$\int_{\partial \phi^2} w = \int_{ \partial \phi} dw =\int_{\phi} d^2 w $$
We know that $d^2$ is zero, so the right most would be integrating zero over a set. If the last equality was true then it would be equal to saying that any possible integral is zero... so that's clearly wrong. So I guess the non sensical part was thinking that my manifold could be written as a boundary of a higher dimensional manifold. Could it be explained what exactly issue with this thinking is? Why can't we think of it in such a way?
 A: If a closed, oriented manifold $M^n$  is oriented boundary of higher dimensional manifold $X^{n+1}$ then it is said to be null-cobordant.
Not all manifolds have this property.
Clearly, all oriented 2-manifolds are null-cobordant and it can be shown that also oriented 3-manifolds are.
However, in dimension 4, it can be shown that $M^4$ is null-cobordant if and only if its signature vanishes $\sigma(M) = 0$ this is due to Rochlin (as a reference see e.g. Scorpan's book the wild world of 4-manifolds).
It is a result of Thom, Pontrjagin and Wall  that a manifold is null-cobordant if and only if all its characteristic numbers (Pontrjagin and Steifel-Whitney) vanish.
This is explained also in Milnor and Stasheef's book on Characteristic classes (see pg 217).
EDIT to answer the comment.
No, you can always find an  $n$-form with non-zero integral on any orientable $n$-manifold, $n>0$ just take its volume form.
What is true is that if $M^n= \partial X^{n+1}$ and  $\omega^n\in \Omega^n(M)$ is  restriction of a form $\hat \omega\in \Omega^n(X)$ which is closed over $X$ i.e. $d\hat\omega = 0$, then you can apply Stokes to conclude that $\int_M \omega = \int_X d\hat \omega = 0$.
More abstractly, there are relative de Rham cohomology groups (see Bott-Tu's book pg. 78) $H^\bullet(X,M)$ and a map $\alpha: H^{n}(M)\to H^{n+1}(X,M)$ (giving the  exact sequence of the pair $(X,M)$).
Then Stokes tells you, that if $\omega^n\in \Omega^n(M)$, is closed in $M$, i.e. $d\omega = 0$, and $\alpha([\omega]) = 0 \in H^{n+1}(X,M)$, then $\int_M \omega = 0$ as well.
In other words, the forms to which you can apply the Stokes argument above, are exactly those in the kernel of $\alpha$.
