Let $R$ be a commutative ring with unity and let $M$ be a finitely generated noetherian $R$-module.
Suppose we are given an $R$-homomorphism: $$ \varphi: R^{(2)}\to R^{(2)} $$ Fixing a basis of $R^{(2)}$ we can obtain $\varphi$ as a $2\times 2$ Matrix with entries in $R$. Let's say it's given as \begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix} This induces the map $$ \varphi_*:hom_R(R^{(2)},M)\to hom_R(R^{(2)},M),\quad f\mapsto f\circ\varphi $$ Now since $hom_R(R^{(2)},M)\simeq M\oplus M$, there should be a map "belonging" to $\varphi_*$ that goes from $M\oplus M$ to $M\oplus M$.
I tried to understand what this map exactly does and following the isomorphism $hom_R(R^{(2)},M)\simeq M\oplus M$ I came to the result that it somehow looks like this: $$ \varphi_*^{'}:M\oplus M \to M\oplus M,\quad (m_1,m_2)\mapsto (m_1,m_2)\left[\begin{array}{l}r_{11}&r_{12}\\r_{21}&r_{22}\end{array}\right]=(m_1r_{11}+m_2r_{21}, m_1r_{12}+m_2r_{22}) $$ Is this correct? And if yes how can I express the Image of that map, because I am having a hard time trying to do this.