$AG $ is idempotent and $W=R(A)\oplus N(AG)\implies V=R(G)+ N(A)$. Let $A\in L(V,W)$. Then the following are equivalent for any linear transformation $G:W\to V$.

$(1)$ $AG $ is idempotent and $W=R(A)\oplus N(AG)$.
$(2)$ $AG$ is idempotent and $V=R(G)+ N(A)$.

For the $(1)\implies (2)$, I guessed that, for any $x\in V$, writing $x=GAx+x-GAx$ will give the answer.
But I am having doubt in proving that
$A(x-GAx)=0$.
Suppose I write like that,
$$Ax-AGAx=Ax-Ax=0$$
what is the wrong in this argument. I am guessing that I can not use $AGAx=Ax$.
Kindly clarify that.

The idea is good. Since $GAx\in R(G)$, you want to see that $y=A(x-GAx)=0$. It's obvious that $y\in R(A)$; however
$$
AGy=AGAx-AGAGAx=AGAx-(AG)^2Ax=AGAx-AGAx=0
$$
so $y\in N(AG)$. As $R(A)\cap N(AG)=\{0\}$, you have the conclusion.
The proof of (2)$\implies$(1) is similar.

 Let $y\in W$; then $y=AGy+(y-AGy)$. Since $AGy\in R(A)$, we want to prove that $y-AGy\in N(AG)$, that is, $AG(y-AGy)=0$, which is easy. We also need to prove that $R(A)\cap N(AG)=\{0\}$, so let $y$ belong to the intersection. Since $y\in R(A)$ we can write $y=A(x+x')$, where $x\in R(G)$ and $x'\in N(A)$: however $Ax'=0$, so $y=Ax$ and $x=Gz$ for some $z\in W$. Since $AGy=0$, we have $0=AGy=AGAx=AGAGz=AGz=Ax=y$.

