Let $M$ be a right $R$-module and $A_1\subseteq A_2\subseteq...$ a chain of direct summands, so there are submodules $B_i$ such that $A_i\oplus B_i=M$.
Is then $B_1\supseteq B_2\supseteq...$? And why or why not?
I know that I can write $A_i$ as $f_i(M)$ for $f_i\in End_R(M)$ with $f_i^2=f_i$ and that then $B_i=(1-f_i)(M)$ but I don't know how to use that.
No, this is typically not true. For instance, let $M=R^3$ and let $A_1$ be the submodule generated by $(1,0,0)$ and $A_2$ be the submodule generated by $(1,0,0)$ and $(0,1,0)$. Then the submodule $B_1$ generated by $(0,1,0)$ and $(0,0,1)$ is a complement of $A_1$ and the submodule $B_2$ generated by $(1,0,1)$ is a complement of $A_2$, but $B_2\not\subseteq B_1$.
However, if you are allowed to choose the $B_i$, you can always arrange for them to form a chain. Starting with a choice of $B_1$ and $B_2$, replace $B_2$ with $B_2'=(1-f_1)(B_2)$. Then $B_2'$ is another complement of $A_2$ (I'll leave this for you to check), and $B_2'\subseteq B_1$. You can then similarly replace $B_3$ with another complement $B_3'$ of $A_3$ such that $B_3'\subseteq B_2'$, and so on.