I'm curious about the general case but also need to understand if $P(A|B)(C|D)$ is the same as $P(A|B) \times P(C|D).$
In equation (7), it states
Is this the same as $P(u|f,s) \times P(f|s)$ ?
The article is available at https://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=2297650


Your understanding is correct. I think there is a typo in the original source. In general it holds that \begin{equation} p(x,y|z) = p(x|y,z) p(y|z) \end{equation} and therefore, \begin{equation} p(x|z) = \sum_y p(x,y|z) = \sum_y p(x|y,z) p(y|z) \end{equation}