Need help to understand a theorem about direct sums and regular morphism of R-modules

I shall work in $R-Mod$, the category of left $R$-modules. Suppose $exists G:Nrightarrow M$ such that $Fcirc Gcirc F=F $.

Consider the short exact sequence$$0 rightarrow operatorname{Ker}Fhookrightarrow Mxrightarrow{F} operatorname {Im}Frightarrow 0$$ Look at $G|_{operatorname{Im}F}:operatorname{Im}Frightarrow M$. We have $Fcirc G|_{operatorname{Im}F}=id_{operatorname{Im}F}$ by the condition. As such the above sequence is a split short exact sequence (it right splits and hence left splits as well). $therefore operatorname{Ker } F$ is a direct summand of $M$.

Consider the short exact sequence $$0rightarrow operatorname{Im}Fxrightarrow{j} N rightarrow operatorname{CoKer}Frightarrow 0$$
Look a $Fcirc G:Nrightarrow operatorname{Im}F$. Once again, we have $Gcirc Fcirc j=id_{operatorname{Im}F}$. So the above sequence left splits and hence is a split short exact sequence. $therefore operatorname{Im}F$ is a direct summand of $N$.

Now assume $operatorname{Ker}F$ is a direct summand of $M$. Then $$0 rightarrow operatorname{Ker}Fhookrightarrow Mxrightarrow{F} operatorname {Im}Frightarrow 0$$ is a split short exact sequence. So we get $g:operatorname {Im}Frightarrow M$ such that $Fcirc g=id_{operatorname{Im}F}$. We also have a sub-module $N'$ of $N$ such that $N=operatorname{Im}Foplus N'$ Define $$G:Nrightarrow M$$ $$(x,y)mapsto g(x)$$

Then $Fcirc Gcirc F(x)=Fcirc G((F(x),0))=Fcirc g(F(x))=F(x) $