We know $m$-th de Rham cohomology group on $U$ is defined to be, $H^{m}_{dR}(U)=ker(d^m)/im(d^{m-1})$ where $d^m:\Omega^m(U)\to \Omega^{m+1}(U)$'s are usual exterior derivative maps. Now its saying $H^{0}_{dR}$ is kernel of map $d:\Omega^{0}(U)\to \Omega^{1}(U)$.
Now my question is to get $H^{0}(U)$ we need $d^{-1}$ defined on $\Omega^{-1}(U)$. But what does $\Omega^{-1}(U)$ mean ?
As a definition, you should take $\Omega^k (U)$ to be zero for all $k$ not in $\{0, 1, \dots, n\}$, where $n$ is the dimension of $U$. So, for example, "$d^{-1}$" is the zero map, and $\text{im} ~ d^{-1} = \{0\} \subset \Omega^0(U)$. Thus $H^0(U) = \text{ker} ~ d_0$ (which consists of locally constant functions). Similarly, $d_n = 0$, so $\text{ker} ~ d_n = \Omega^n(U)$, and $H^n(U) = \Omega^n(U)/\text{im} ~ d_{n-1}$.