For Fubini's theorem to hold, we must have $f\in L^1(\mathbb R^2)$, that is
$$\iint |f(x,y)|\ \mathsf d(x\times y)<\infty. $$
But here we have
\begin{align}
\iint |f(x,y)|\ \mathsf d(x\times y) &= \iint|\chi_{[0,\infty]\times[x,x+1)}(x,y)-\chi_{[0,\infty]\times [x+1,x+2)}(x,y)|\ \mathsf d(x\times y)
\end{align}
Since $[0,\infty]\times[x,x+1)$ and $[0,\infty]\times[x+1,x+2)$ are disjoint for all $x$, the above integral is equivalent to
$$\iint\chi_{[0,\infty]\times[x,x+1)}(x,y)\ \mathsf d(x\times y) - \iint\chi_{[0,\infty]\times[x,x+1)}(x,y)\ \mathsf d(x\times y), $$
and both of these integrals are infinite, so this is of the form $\infty-\infty$ and $f\notin L^1(\mathbb R^2)$. As for the iterated integrals, we have
\begin{align}
\iint f\ \mathsf dy\ \mathsf dx &= \int_0^\infty\int_0^\infty \left(\chi_{[x,x+1]}(y)-\chi_{[x+1,x+2]}(y)\right)\ \mathsf dy\ \mathsf dx\\
&= \int_0^\infty\left[\int_x^{x+1}\ \mathsf dy - \int_{x+1}^{x+2} \ \mathsf dy \right]\mathsf dx\\
&= \int_0^\infty 0\ \mathsf dx\\
&=0
\end{align}
and integrating in the opposite order yields some finite number $c$ (for $x,y<2$) plus
\begin{align}
\int_2^\infty \left[ \int_{y-1}^y\ \mathsf dy - \int_{y+1}^y\ \mathsf dt\right]\mathsf dy &=
\int_2^\infty (x-(x-1) -(x+1)+x)\ \mathsf dx\\
&= \int_2^\infty 0\ \mathsf dx\\
&= 0.
\end{align}
The value of the integral is then
\begin{align}
c &= \iint\limits_{[0,2]\times [0,2]}f(x,y)\ \mathsf d(x\times y)\\
&= \iint (\chi_{[0,y)\times[0,1]}(x,y) + \chi_{[1,y-1]\times[1,2]}(x,y) - \chi_{(0,y]\times[1,2]}(x,y) +\chi_{[1,y]\times[1,2]}(x,y))\ \mathsf d(x\times y)\\
&= \frac12 + \frac12 - \frac12 + \frac12\\
&= \frac32.
\end{align}
Best Answer
Fubini's theorem requires the integrated function to be integrable in the product space. Tonelli's theorem only requires non-negativity. The combined result is often referred to as Fubini-Tonelli. In your case, you are using the Tonelli part of the theorem.
Let $x>0$. Substituting $z=x-y$ in the inner integral gives
$$F(x)=\int_{-\infty}^\infty |2(x-y)e^{-(x-y)^2}| dy=\int_{-\infty}^\infty 2|z|e^{-z^2}dz=2$$
Therefore $F(x)=2$ for every $x>0$. Thus your integral equals
$$\int_{\mathbb{R}^2} |f(x,y)| d(x,y)=\int_0^\infty 2\, dx = \infty$$
That means your function $f$ is not integrable in the product space $\mathbb{R}\times \mathbb{R}$.