We know that in differential geometry, $d^2\omega=0$, where $\omega $ is a form and $d$ is the exterior derivative.
However if this form happens to be the exterior derivative of another form $\theta$ such that $\omega =d\theta $ then won't $d\omega $ always be zero since we could also write $d\omega =d(d\theta)$?
Forgive me if I'm missing something elementary!
That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed.
However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare lemma says that any closed form is locally exact.) There is a gadget called the de Rham cohomology (of the manifold that we are working on) that measures this failure, and it is an excellent tool that one can use to study the topology of smooth manifolds.
That's precisely correct! See This wikipedia page on closed and exact forms for more details.
