If I understand the question correctly, then here is a likely answer.
But before getting there, let me say that this is a very poorly formed question. If you are asking for help, then put at least as much effort into writing your question as the people who respond put into their answer.

For any line bundle $L$ on $Z$, and a global section of $L^{\otimes m}$ there is an associated cyclic cover of $Z$ of degree $m$ branched over exactly the zero locus of the given section. Furthermore, the pull-back of the given section to this cover is the $m^\text{th}$ power of a section of the pull-back of $L$.
(For more on this, see for example page 67 of this book.)

A section of $f_*(K_{X/B}^{\otimes m})$ could be considered a family of global sections of $K_{X_b}^{\otimes m}$ and looking at the construction of the cyclic cover it is relatively easy to see that these covers form a family.

From what you are saying I assume that there might be an additional assumption on $K_{X/B}$, maybe something like it is relatively semi-ample or something like that.

The point is that if you $K_{X/B}^{\otimes m}$ is relatively generated by global sections, then choosing a general section will give you something smooth, at least globally, so the zero locus on the fiber is smooth at least for a general $b\in B$. Then performing the cyclic cover gives another smooth variety (and you need to have some control of the singularities if you want to talk about canonical forms on the cover).
Now if the cover is indeed smooth, then by Hurwitz's theorem the pull-back of $K_{X/B}$ to the family of the covers embeds into the relative canonical of that family. By the above general result about the section pulling back to an $m^\text{th}$ power on the family of covers gives you the section of the relative canonical of the family of covers which then can be considered as a family of canonical forms on these covers.

divisorto be relatively semi-ample. Your question talks about amorphismbeing relatively semi-ample. I think I will withdraw from this question now. $\endgroup$2more comments