If $e$ and $f$ are idempotents, in a ring $R$. My question is how I can show $Re+Rf=Re+R(f-fe)$.
You have to prove that $Re+Rf \subseteq Re+R(f-fe)$ and $Re+Rf \supseteq Re+R(f-fe)$.
The second inclusion is clear because $xe+y(f-fe) = (x-yf)e+yf \in Re+Rf $.
For the first inclusion, it suffices to prove that $f \in Re+R(f-fe)$. But this is easy because $Re+R(f-fe) \ni fe+f(f-fe)=f$. In other words, $xe+yf = (x+yf)e+(yf)(f-fe)$.
I don't think that $e$ has to be idempotent.