The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted.
The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the deviation of the actual force from that average.
Why do we do it that way? When looking at the tidal forces on the oceans we treat the Earth as a rigid body with spherical symmetry. With that, we can use a variation of Newton's Shell theorem to say that the extended Earth will move the same way as if it were a point mass at its center.
Now the oceans are fluid (the opposite of rigid) and each bit responds to the Moon's force locally.
That force is (skipping constants unnecessary to make the cartoon plot)
$$F = -\frac{\mathbf{\hat{r}}}{|r|^2} = -\frac{\mathbf{r}}{|r|^3}$$
where the vector $\mathbf{r}$ is drawn from the Moon to some point on the Earth and $\mathbf{\hat{r}}$ is its unit vector. If the Earth's center is at $\mathbf{\hat{x}} R$ ($R$ is the Moon-Earth distance) and you subtract $-\mathbf{\hat{x}}/R^2$ you'll get that image.
In the plot below I've chosen the Earth-Moon distance to be only 10 Earth radii to highlight the slight left-right asymmetry. The tidal force is stronger on the side closer to the Moon.
import numpy as np
import matplotlib.pyplot as plt
R = 10.0
r_moon = np.array([R, 0], dtype=float)[:, None]
earth = np.zeros(2)[:, None]
theta = np.linspace(0, 2*np.pi, 49)
positions = earth + np.array([f(theta) for f in (np.cos, np.sin)])
r = positions-r_moon
F = -r * ((r**2).sum(axis=0))**-1.5
r = earth-r_moon
Fmean = -r * ((r**2).sum(axis=0))**-1.5
Ftide = F - Fmean
if True:
plt.figure()
plt.subplot(2, 1, 1)
(x, y), (Fx, Fy) = positions, 50.*Ftide
plt.quiver(x, y, Fx, Fy, width=0.005)
plt.plot(x, y, '-b')
plt.xlim(-2, 2)
plt.ylim(-1.5, 1.5)
plt.gca().set_aspect('equal')
plt.subplot(4, 1, 3)
for thing in F:
plt.plot(thing)
plt.subplot(4, 1, 4)
for thing in Ftide:
plt.plot(thing)
plt.show()
