In this context $R$ is a finite dimensional algebra over a field $K$ and $U$ is a finite dimensional $R$-module. It goes $U/(\text{Rad}R \cdot U)$ is a module over $R/\text{Rad}(R) $ which is a semisimple algebra. Hence $U/(\text{Rad}R \cdot U ) $ is a semisimple module.
Why is this the case? I take it this must be some sort of general result about semisimple algebras?
If you are asking why $U/(\text{Rad}R \cdot U )$ is a semisimple $R/\text{Rad}R$ module, then you can see why all modules of a semisimple ring are semisimple at this post.
If you are asking why it is a semisimple $R$ module, then that is because $R$ and $R/\mathrm{Rad}R$ have "the same simple modules." Something is a direct sum of simple $R$ modules iff it is a direct sum of simple $R/\mathrm{Rad}(R)$ modules.
(Throughout I assume $\mathrm{Rad}(R)$ means the Jacobson radical, which I'd normally denote $J(R)$. $\mathrm{Rad}$ is a bit more overloaded.)
