ring homorphism from $\mathbb{Z+Z}$ to $\mathbb{Z+Z}$ Determine all the ring homorphisms from $\mathbb{Z+Z}$ to $\mathbb{Z+Z}$

This homorphism can be determined by $f((0,1))$ and $f((1,0))$ then how to proceed further?
This question is discussed here Determine all ring homomorphism from$ \mathbb{Z}\oplus \mathbb{Z}$into $\mathbb{Z}\oplus \mathbb{Z}$
But I am not understanding.
Note that $(0,1)$ and $(1,0)$ both square to themselves: $$(1,0)(1,0)=(1,0),\qquad (0,1)(0,1)=(0,1).$$  Thus they must map to elements of the form: $(x,y)$ with $$(x,y)=(x,y)(x,y)=(x^2,y^2).$$  Thus $x^2=x$ and $y^2=y$ so $x,y\in\{0,1\}.$
That gives $16$ possibilities, with each of $(1,0)$ and $(0,1)$ mapping to one of $(0,1),(1,0), (1,1), (0,0)$.
However a ring homomorphism must map $(1,1)$ to $(1,1)$.  Thus you are left with $4$ ring homomorphisms:
$$\begin{array}{ccccccc}(1,0)\mapsto (0,0), &amp;&amp; (1,0)\mapsto (0,1),&amp;&amp;(1,0)\mapsto (1,0),&amp;&amp;(1,0)\mapsto (1,1).\\
(0,1)\mapsto (1,1) &amp;&amp; (0,1)\mapsto (1,0)&amp;&amp;(0,1)\mapsto (0,1)&amp;&amp;(0,1)\mapsto (0,0)
\end{array}$$
Note these correspond to the identity, swapping the factors, and projecting the first or second factor onto the diagonal.
Well, $f(a,b) = a\cdot f(1,0) + b\cdot f(0,1)$, where $f(1,0), f(0,1)\in{\Bbb Z}^2$ are fixed.