Maybe this question is rather obvious but I didn't manage to solve it myself.
Assume $M$ is a closed, oriented manifold. take $$ \Omega^k(M)\ni \omega = \begin{cases} d\beta~,~~~ in~ U\\ 0~,~~~ otherwise \end{cases} $$ Where $U\subset M$ is open, and $\beta \in \Omega^{k-1}(U)$. $\omega$ has a compact support in $U$.
Then $\omega$ is an exact form?
On
COMPLETE EDIT: Now that you have removed cohomology and the question has been clarified in my mind, I believe you are asking the following question.
Suppose $\omega$ is a $k$-form with compact support contained in $U$ and there is a smooth form $\beta$ on U with $d\beta=\omega$ on $U$; is $\omega$ globally exact on $M$?
The answer is NO, not necessarily. Consider $M=S^1$, and let $U=S^1-{(1,0)} \cong \Bbb R$. On $\Bbb R$ take a compactly supported function $f$ with integral $1$, and consider $\omega = f(x)\,dx$. This extends to be a smooth $1$-form on all of $S^1$, since $\omega$ is identically $0$ outside a compact subset of $\Bbb R$. $\int_{S^1}\omega = 1$, so $\omega$ cannot be exact. However, on $U=\Bbb R$, we can set $\beta(x) = \int_0^x f(t)\,dt$. By the Fundamental Theorem of Calculus, $d\beta = \omega$ on $U$.
If it is $\beta$ that has compact support in $U$, then it is true. Denote support of $\beta$ by $K$, obviously $U^c$ is a closed set that is disjoint from $K$. Then there exists a separating function $f$ on $M$ taking the value in $[0,1]$, with $f(K)\equiv1$ and $f(U^c)\equiv0$. Then, $$ d(f\beta)=df\wedge\beta+fd\beta=\begin{cases} d\beta & \text{on }U \\ 0 & \text{on }U^c \end{cases} $$ and therefore $d(f\beta)=\omega$ on the whole of $M$ and $\omega$ is exact.