Here's the section of the plane around the square, color-coded by which point the flea will jump over:
$\hskip2in$
(The $x$ axis points rightwards and the $y$ axis points upward.)
We can model the process mathematically as
$$f(x, y) = \begin{cases}
(\phantom{-}2-x,\; \phantom{-}2-y) & \quad \color{red}{x \geq 1\phantom{-} ~\text{ and }~ y < 1\phantom{-}} \\
(\phantom{-}2-x,\; - 2-y) & \quad \color{orange}{x < 1\phantom{-} ~\text{ and }~ y \leq -1} \\
( - 2-x,\; - 2-y) & \quad \color{blue}{x \leq -1 ~\text{ and }~ y > -1} \\
( - 2-x,\; \phantom{-}2-y) & \quad \color{green}{x > -1 ~\text{ and }~ y \geq 1\phantom{-}}
\end{cases}$$
where the first case is a reflection over point ${A(1, 1)}$ if $(x, y)$ is in the red part of the plane; the second case is a reflection over point ${B(1, -1)}$ if $(x, y)$ is in the yellow part of the plane, etc.
Through lots of case splitting, we can find a similar description of $f^2$. That is, we can divide the entire plane (minus the square) into eight regions, and specify which transformation will apply to the flea's coordinates after jumping twice from its current position. They are:
$\hskip2in$
$$f^2(x,y) = \begin{cases}
(x-4, y-4) & \quad \color{red}{ x \geq 3 ~\text{ and }~ y < 1} \quad \text{(region 1)}\\
(x-4, y\color{white}{+0}) & \quad \color{red}{ x \in [1, 3) ~\text{ and }~ y < 1} \quad \text{(region 2)}\\
(x-4, y+4) & \quad \color{orange}{ x < 1 ~\text{ and }~ y \leq -3} \quad \text{(region 3)}\\
(x\color{white}{+0}, y+4) & \quad \color{orange}{ x < 1 ~\text{ and }~ y \in (-3, -1]} \quad \text{(region 4)}\\
(x+4, y+4) & \quad \color{blue}{ x \leq -3 ~\text{ and }~ y \geq -1} \quad \text{(region 5)}\\
(x+4, y\color{white}{+0}) & \quad \color{blue}{ x \in (-3, -1] ~\text{ and }~ y \geq -1} \quad \text{(region 6)}\\
(x+4, y-4) & \quad \color{green}{ x > -1 ~\text{ and }~ y \geq -3} \quad \text{(region 7)}\\
(x\color{white}{+0}, y-4) & \quad \color{green}{ x > -1 ~\text{ and }~ y \in (-3, -1]}\quad \text{(region 8)}
\end{cases}$$
We will prove that making repeated double jumps starting from $(4k, 1)$, where $k$ is a positive integer, leads us to $(4k+4, 1)$. Specifically, this will happen after $4k+4$ double jumps (or $8k+8$ jumps in total.)
- First, $(4k, 1)$ is (right on the boundary of) region 8.
- After one region 8 double jump (south), we are in $(4k, -3)$, in region 1.
- $k$ region 1 double jumps (south-west) later, we end up in $(0, -3-4k)$, in region 3.
- $k+1$ region 3 double jumps (north-west) later, we end up in $(-4k-4, 1)$, in region 5.
- $k+1$ region 5 double jumps (north-east) later, we end up in $(0, 4k+5)$, in region 7.
- $k+1$ region 7 double jumps (south-east) later, we end up in $(4k+4, 1)$, back on the boundary of region 8.
The final path looks like this (in purple):
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In total we made $4k+4$ double jumps in a diamond-shaped path around the origin. (Under the taxicab geometry, the first step is the one that actually brings us farther from the origin, while the others don't change the taxicab distance from the origin at all.)
Thus, we can start at $(4, 1)$, make 16 jumps to end up at $(8, 1)$, make 20 more jumps to end up at $(12, 1)$, etc., and go as far as we like. Specifically, we reach $(124, 1)$ after 3960 jumps, where the ratio of distances is
$$\frac{\sqrt{124^2+1}}{\sqrt{4^2+1}} = 30.075395... > 30.$$