On page 27 of that book, it is claimed that the inclusion map $\delta$ which maps a form from the non-empty intersection of two open covers of the circle to the disjoint union of those covers has a one dimensional kernel:
$$ \delta : H_c^1 (U\cap V) \rightarrow H_c^1 (U \sqcup V) $$ $$ w= (w_1 , w_2) \rightarrow (-(j_U)_* w , (j_V)_* w)$$
where inclusion simply extends the form by zero outside the intersection of U and V. Clearly, $(-(j_U)_* w , (j_V)_* w)$ is zero only if w is zero. So Kernel of $\delta$ is not one dimensional. Mistake? note the $U\cap V$ is two disjoint "arcs".
It is not true that $(-(j_U)_* \omega , (j_V)_*\omega)=0$ iff $\omega=0$. Note that $(j_U)_*\omega$ and $(j_V)_*\omega$ here are cohomology classes in $H^1_c(U)$ and $H^1_c(V)$, not just $1$-forms. So we have to consider the possibility that they might be coboundaries. A $1$-form on $U$ is a coboundary iff its integral is $0$, and similarly for $V$. So $\ker(\delta)$ consists of exactly those $\omega$ such that $\int_{U\cap V}\omega=0$. Since there is an isomorphism $H^1_c(U\cap V)\to \mathbb{R}\oplus \mathbb{R}$ given by integration over each of the two components of $U\cap V$, $\ker(\delta)$ corresponds to the one-dimensional subspace $\{(x,-x):x\in\mathbb{R})\}\subset\mathbb{R}\oplus\mathbb{R}$.