When one is exactly at the critical angle, the light behaves in a way that may be interpreted as "something in between" refraction and reflection: it continues in a direction that is tangent to the boundary of the mediums.
When the angle is smaller than the critical angle, we get refraction. At the critical angle, $\theta_2$ of the refraction becomes 90 degrees, so we get the tangent propagation. At angles larger than the critical ones, there is a discontinuity: the equation for $\theta_2$ (arcsine of something) has no solutions which is why we get a total internal reflection.
There is nothing about these facts that would contradict reversibility or time-reversal symmetry of the laws of physics. If we time reverse the behavior at the critical angle, it indeed looks like the light must "randomly pick" a moment at which it enters the medium with the higher refraction index and there isn't any unique way to pick the preferred moment.
But that's not a problem because the probability that the direction of light is "exactly" tangent to the boundary is zero. In a real-world situation, the light will be a superposition of beams with angles $\theta_2=\pi/2-\epsilon$ for various small values of $\epsilon$, and for any nonzero $\epsilon$, the light will know very well when it hits the boundary. So your problem only occurs at a negligible, "measure zero" portion of the situation, so it is at most a "measure zero" problem. When one adds the appropriate degree of realism and specifies the precise angles and deviations from the "idealized model", the problem goes away.
The situation shown in the figure pertains to the situation where the polarizer and the analyzer having their "pass-axis" aligned.
Since any one of them shall eliminate light/electric-field component along the axis perpendicular to the pass-axis, if linearly polarized light is incident onto a polarizer, then the output of the polarizer shall have light with electric field oscillations only along one definite axis, and nothing along the perpendicular axis. Now, if the pass-axis of the analyzer is also aligned with this axis, output light of the analyzer shall be the entire light input onto the analyzer, since no component is eliminated. That is to say, this would be the position of maximum light intensity through the analyzer.
However, if you rotate the pass-axis of the analyzer about the pass-axis of the polarizer through an angle $\theta$, only the $\cos \theta$ component of the light shall be parallel to the pass axis, and shall pass through, while the other $\sin \theta$ component would be blocked by the analyzer, being perpendicular. Thus, the intensity in this position would be $I = I_0 \cos^2 \theta$, where $I_0$ is the maximum light intensity. (Malus Law.)
If you increase $\theta$, moving it completely between $0$ and $2\pi$, i.e. full rotation in a circle, there will be two points $\theta = \pi/2$ and $3\pi/2$, where light oscillations shall be exactly perpendicular to the pass axis, and hence observed intensity would vanish. On the other hand, there are also two positions $\theta = 0$ and $\pi$, where $I = I_0$ (maximum).
This is how you analyze linearly polarized light using these instruments. Also, if the nature of the original light is unknown, this is how you learn that the incident light was originally linearly polarized.
Best Answer
You are right. The reflected beam is always there. It is just that when incident angle becomes equal to the critical one, the intensity of the reflected beam becomes equal to the incident one, and then stays that way as you increase the incident angle further.
The choice to not to draw the reflected beam at the critical incident angle is purely a figure design issue. Probably to not to overcrowd the figure.