The function is convex. First write the function as
$$g(x_1, x_2) = \cases{
-x_1-x_2+2\alpha+\beta & $x_1 \leq \alpha$, $x_2 \leq \alpha+\beta$\\
-x_1+x_2-\beta & $x_1 \leq \alpha$, $x_2 \geq \alpha+\beta$\\
2x_1-x_2-\alpha+\beta & $x_1 \geq \alpha$, $x_2 \leq x_1+\beta$\\
x_2-\alpha+\beta & $x_1 \geq \alpha$, $x_2 \geq x_1+\beta$}$$
Note that $g$ is continuous (by its original definition) and piecewise linear, in that we can partition its domain into finitely many (closed, convex) regions, such that it is affine on each region.
Let's take a diversion to find a condition we'll use to verify that $g$ is convex. A function $f : D \to \mathbb{R}$ defined on a convex domain $D \subset \mathbb{R}^d$ is convex iff the function $f_{x, y} : [0, 1] \to \mathbb{R}$ given by $f_{x, y}(t) = f((1-t)x + ty)$ is convex for any $x, y \in D$. But when $f$ is piecewise linear, so is any $f_{x, y}$, and in this case it is easy to prove that $f_{x, y}$ is convex iff its sequence of slopes (one for each successive region of its domain) is increasing. Say that $f$ has regions $R_1, \dots, R_n$, where $f(u) = v_i \cdot u + c_i$ on $R_i$. Then if the line $L_{x, y} = \{(1-t)x + ty \mid t \in [0, 1]\}$ intersects the successive regions $R_{i_1}, \dots, R_{i_k}$ (going from $t = 0$ to $t = 1$), then the sequence of slopes is $v_{i_1} \cdot (y-x), \dots, v_{i_k} \cdot (y-x)$, so we must check that this sequence is increasing for any $x, y$.
To check this, we can equivalently check the following simpler condition: that whenever $x' \in R_i$ and $y' \in R_j$, we have $v_i \cdot (y'-x') \leq v_j \cdot (y'-x')$. This condition is clearly necessary, and is sufficient since then for any $1 \leq r < s \leq k$, taking $x' = (1-t_1)x + t_1y \in R_{i_r}$ and $y' = (1-t_2)x + t_2y \in R_{i_s}$ (where $t_1 < t_2$), we have that $v_{i_r} \cdot (y' - x') \leq v_{i_s} \cdot (y'-x')$, i.e. that $(t_2 - t_1)(v_{i_r} \cdot (y-x)) \leq (t_2 - t_1)(v_{i_s} \cdot (y-x))$ hence $v_{i_r} \cdot (y-x) \leq v_{i_s} \cdot (y-x)$, meaning the sequence of slopes for any $L_{x, y}$ is increasing as desired. But $v_i \cdot (y-x) \leq v_j \cdot (y-x)$ iff $(v_j - v_i) \cdot x \leq (v_j - v_i) \cdot y$, so we can restate our simpler condition as the following result:
$f$ is convex if and only if for every $R_i, R_j$, we have $\sup_{x \in R_i} (v_j - v_i) \cdot x \leq \inf_{x \in R_j} (v_j - v_i) \cdot x$.
and in fact this only needs to be checked for neighboring $R_i, R_j$, namely those with $R_i \cap R_j \neq \varnothing$.
We can now use this to check that $g$ is convex. Label the regions $R_1, R_2, R_3, R_4$ in the same order as listed above, so $v_1 = (-1, -1), v_2 = (-1, 1), v_3 = (2, -1), v_4 = (0, 1)$. Checking, we see that
\begin{align*}
\sup_{x \in R_1} (v_2 - v_1) \cdot x &= 2\alpha+2\beta &= \inf_{x \in R_2} (v_2 - v_1) \cdot x \\
\sup_{x \in R_1} (v_3 - v_1) \cdot x &= 3\alpha &= \inf_{x \in R_3} (v_3 - v_1) \cdot x \\
\sup_{x \in R_1} (v_4 - v_1) \cdot x &= 3\alpha+2\beta &= \inf_{x \in R_4} (v_4 - v_1) \cdot x \\
\sup_{x \in R_2} (v_3 - v_2) \cdot x &= \alpha-2\beta &= \inf_{x \in R_3} (v_3 - v_2) \cdot x \\
\sup_{x \in R_2} (v_4 - v_2) \cdot x &= \alpha &= \inf_{x \in R_4} (v_4 - v_2) \cdot x \\
\sup_{x \in R_3} (v_4 - v_3) \cdot x &= 2\beta &= \inf_{x \in R_4} (v_4 - v_3) \cdot x
\end{align*}
so $g$ is indeed convex.
