I cannot understand $\tau_{see}$ means in the question

Question

The question is

Time dilation implies that if a clock moves relative to a frame S, careful measurements made by observers in S [as in Fig. 1.22(a), for example] will find that the clock runs slow. This is not at all the same thing as saying that a single observer in S will see the clock running slow; and the latter statement is, in fact, not always true. To understand this, remember that what we see is determined by the light as it arrives at our eyes. Consider the observer Q in Fig. 1.22(b) and suppose that as the clock moves from A to B, it registers the passage of a time $\tau_0$. As measured in S, the time between these two events (“clock at A” and “clock at B”) is of course $\tau$ = $\gamma$ $\tau_0$.However, B is closer to Q than A is; thus light from the clock when at B will reach Q in a shorter time than will light from the clock when at A. Therefore, the time $\tau_{see}$ between Q’s seeing the clock at A and seeing it at B is less than $\tau$

Answer 1

τ_see is the time Q records between the clock reaching A and the clock reaching B.

Since photons take time to travel, the clock is already far past A by the time Q sees it. B is closer to Q so the photons have less distance to travel; by the time Q sees the clock pass B it is a little bit past B. Then the time τ_see is less than τ recorded by the two stationary people since the clock actually doesn't move as far.

Imagine someone coming at you on a bike throws a baseball: by the time you catch the ball they are a few meters closer. A little later, they throw another one, and you catch it a little sooner than you did the first because they are closer. The baseballs may have been thrown 10s apart, but you caught them 8s apart. That is τ_see.