Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and $f\left(x+f(x+2y)\right)=f(2x)+f(2y)$

Supposing that we already know that $f(2n) = 2n + 2$ for $n \geq 0$, we can proceed as follows.

For some positive integer $k$, let $x = 2k$ and $y = -k$ in the functional equation. We get that $$ f(2k + 2) = f( 2k + f(2k - 2k) ) = f(4k) + f(-2k). $$

Now $2k + 2$ and $4k$ are positive even integers, and so this tells us that $$ 2k + 4 = 4k + 2 + f(-2k), $$ and hence $$ f(-2k) = -2k + 2. $$

Thus $f(2n) = 2n + 2$ holds for all integers $n$.

We know show that $f$ sends odd integers to odd integers. Suppose to the contrary that there are some integers $a$ and $b$ such that $$ f(2a - 1) = 2b. $$

Take $x = -1$ in the functional equation to get that for any integer $y$, that $$ f(f(2y - 1) - 1) = f(-2) + f(2y) = 2y + 2. $$

This gives us that $$ \begin{align*} f(2a - 1) & = 2b & f(2b - 1) & = f(f(2a - 1) - 1) = 2a + 2 \\ f(2a + 1) & = f(f(2b - 1) - 1) = 2b + 2 & f(2b + 1) & = f(f(2(a+1)-1)-1) = 2a + 4 \\ f(2a + 3) & = f(f(2(b+1)-1)-1) = 2b + 4 & f(2b + 3) & = f(f(2(a+2)-1)-1) = 2a + 6 \\ & \vdots & & \vdots \end{align*} $$ and inductively, we obtain that for every non-negative integer $k$, we have that $$ f(2a + 2k - 1) = 2b + 2k \quad\text{ and }\quad f(2b + 2k - 1) = 2a + 2k + 2. $$

Now suppose that $b \leq a$. Then there is some $k \geq 0$ such that $a = b + k$, which gives us that $$ 2b = f(2a - 1) = f(2b + 2k - 1) = 2a + 2k + 2 = 2b + 4k + 2, $$ which implies that $4k + 2 = 0$, a contradiction. Similarly, if $b > a$, then there is some $k \geq 0$ such that $b = a + k$. We then obtain that $$ 2a + 2 = f(2b - 1) = f(2a + 2k - 1) = 2b + 2k = 2a + 4k, $$ which gives us that $4k = 2$, again a contradiction.

We conclude that $f(2y - 1)$ is odd for every integer $y$. Thus for every integer $y$, we have that $f(2y - 1) - 1$ is even, and hence we have that $$ 2y + 2 = f(f(2y - 1) - 1) = f(2y - 1) + 1 $$ and hence that $$ f(2y - 1) = 2y + 1 $$ for every integer $y$.

We see that $f(n) = n + 2$ for every integer $n$, and can verify that this does indeed solve the functional equation.

Using your results, we find that $$\tag1f(x+f(x+2y))=2x+2y+4\qquad\text{for }x,y\ge0 $$ In particular, $$\tag2 f(x+f(x))=2x+4\qquad\text{for }x\ge0$$

Let $S=\{\,k\in\Bbb Z\mid f(2k)=2k+2\,\}$. You essentially showed that $k\in S\implies k+1\in S$ and hence from the given $0\in S$, we have $\Bbb N_0\subseteq S$. With $x=-2y$ we have $$ \tag3f(-2y+2)=f(-4y)+f(2y)\qquad\text{for }y\in\Bbb Z$$ Thus if two of $1-y,-2y,y$ are in $S$, then tso is the third. In particular, for $y<0$ we already known $1-x,-2y\in S$; we conclude $S=\Bbb Z$, i.e., $$\tag4f(x)=x+2\qquad \text{for }x\in2\Bbb Z$$ and hence $$\tag{1'}f(x+f(x+2y))=2x+2y+4\qquad\text{for }x,y\in\Bbb Z $$ and in particular $$\tag{2'} f(x+f(x))=2x+4\qquad\text{for }x\in\Bbb Z$$

Assume that for some odd $x=2n+1$ the value $f(x)=2m$ is even. Then for $k\in\Bbb Z$ $$\begin{align}f(1+2k)&=f\bigl(1+2(k-m)+f(2n+1)\bigr)\\ &=f\Bigl((1+2k-2m)+f\bigl((1+2k-2m)+2(n-k+m)\bigr)\Bigr)\\ &= 2(1+2k-2m)+2(n-k+m)+4\\ &=6+2k-2m+2n\end{align}$$ and so with the odd constant $c:=2n-2m+5$ $$f(x)=x+c\qquad\text{for }x\in 2\Bbb Z+1$$ Then $6=f(1+f(1))=f(2+c)=2+2c$ implies $c=2$, contradicting that $c$ is odd. Therefore $f(x)$ is odd for all odd $x$. But then $x+f(x)$ is even and from $(4)$, we get $2x+4=f(x+f(x))=x+f(x)+2$ and so $$f(x)=x+2 $$