Proof of a left R-module M is free with basis X iff M is internal direct summand of Rx where x is in X

Question

I have a confusion in the proof. Let $M$ is internal direct sum of $Rx$ where $x$ is in $X$. I have no trouble in proving that $X$ generates $M$. Now when we show that $X$ is linearly independent.we take distinct elements $y_1,y_2,...y_n$ in $X$ such that $r_1y_1+r_2y_2+...r_ny_n=0$ where $r_i$ is in $R$ for each $i$ from 1 to $n$. Now we say that since $M$ is internal direct sum of $Rx$ where $x$ is in $X$ therefore 0 has a unique representation and hence each $r_i=0$. This is exactly the step I didn't understand, since 0 has unique representation, then I think it should have been said that each $r_iy_i=0$. So please explain why is it $r_i=0$ instead of $r_iy_i=0$.