When considering a continuous $\omega\in\Lambda^1(U)$, where $U\subset\mathbb{R}^n$ is some open connected subset, $d\omega=0$ is insufficient to guarantee exactness since Poincare's Lemma holds for contractible manifolds, and $U$ may not be one.
(Note in below discussion I assume all paths/loops are $\mathcal{C}^1$ singular 1-cells) If every loop $\gamma$ in $U$ satisfies $\int_\gamma\omega=0$, then we know that for any two points $\textbf{a}, \textbf{b}\in U$ any two paths $\gamma_1,\gamma_2\in U$ from $\textbf{a}$ to $\textbf{b}$ can be concatenated such that $\gamma_1-\gamma_2$ forms a loop so we have path independence for every path in $U$ since $\int_{\gamma_1-\gamma_2}\omega =0$.
Complete path independence in $U$ results in exactness since it ensures that $\forall\textbf{a},\textbf{x}\in U$, $f(\textbf{x})=\int_{[\textbf{a},\textbf{x}]}\omega$ is well-defined, which we can use to show $d f=\omega$ in an open neighborhood of every $\textbf{x}$ by looking at the difference of $\lim_{h\rightarrow 0}f(\textbf{x}+h{\boldsymbol\delta}_i)-f(\textbf{x})$ for $1\le i \le n$.
However, this question's answer by @Seub (see "Lemma 3") seems to imply a sufficient condition for exactness is path independence solely about the 2 paths of the oriented cell $\Delta(\textbf{a},\textbf{x})$ (in $U\subset\mathbb{R}^2$). How is this the case? It make sense, looking at how the proof for exactness works given that $\omega$ is completely path independent, but the first statement in that proof ($f$ is well-defined) is now in question.
If, as @Seub mentions, we restrict all possible paths $[\textbf{a},\textbf{x}]$ to just ones on $\partial \Delta(\textbf{a},\textbf{x})$, then I agree $f$ is well defined, and $df=\omega$ on the boundary of $\Delta(\textbf{a},\textbf{x})$. Why can we then "extend" its exactness to an open interval around $\textbf{b}$? Does it have something to do with continuity?
