From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $\mathbb{C}$ which is not all of $\mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.
Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?
It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U \to U_1(0)$ extends to homeomorphism $\bar{h} : \overline{U} \to \overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.