The clocks are then synchronized to time $t_0$ [...]
Which clocks, and in which frame? You must always remember that synchronization of clocks which are sitting at different points in space is a frame-dependent property. I assume that you mean that clocks $A$ and $B$ are synchronized in the rest frame at $A$ (or $B$), and that the clock on the spaceship is synchronized with clock $A$ as it passes by. However, note that in this case the clock at $B$ will not be synchronized with the others in the rest frame of the ship.
When spaceship meets B, the first scenario shows the rod clock showing larger time value than the spaceship clock, while the the second scenario shows the opposite. Both cannot be true, because one must occur when B and spaceship meet.
The $t_2$ to which you refer in your first scenario is different from the $t_2$ in your second scenario. In the first case, according to the clock at $B$, the ship arrives at $B$ at time $t_2 = L/u$, where $L$ is the distance between $A$ and $B$ in the rest frame of the observer at $B$. The ship's clock reads $T = t_2/\gamma = L/u\gamma$, where $\gamma = (1-u^2/c^2)^{-1/2}$.
In the second scenario, an observer on the ship sees the rod length contracted, with apparent length $L' = L/\gamma$. Accordingly, point $B$ reaches the ship when the ship's clock reads $T=L'/u = L/u\gamma$, in agreement with the previous calculation.
Furthermore, in the second scenario the clock at $B$ was never synchronized with the clock on the ship. Let $(x',t')$ be the coordinates in the ship's frame and $(x,t)$ be the coordinates in the Earth frame centered at clock $A$. The event $(x,t)=(x',t')=(0,0)$ corresponds to the ship passing clock $A$. According to the Lorentz transform equations,
$$t' = \gamma\left(t-\frac{ux}{c^2}\right)$$
When the ship passes clock $A$, then the ship's clock reads $t'=0$, and clock $A$ (which sits at $x=0$) also reads $t=0$. However, clock $B$ is sitting at $x=L$, which means that
$$t - \frac{uL}{c^2} = 0 \implies t = \frac{uL}{c^2}$$
In other words, clock $B$ starts with an offset of $uL/c^2$ according to the observer on the ship.
With that in mind, the relativist riding on the ship would recognize that when she reaches clock $B$, then it would read
$$t_2 =T/\gamma + \frac{uL}{c^2}=\frac{L}{u\gamma^2} + \frac{uL}{c^2} = \frac{L}{u}\left(\frac{1}{\gamma^2} +\frac{u^2}{c^2}\right) = \frac{L}{u}$$
where the first term is the reading on the ship's clock divided by $\gamma$ (due to the time dilation) and the second term is the initial offset on clock $B$. This is fully in agreement with our previous calculations, as expected.
"However, note that in this case the clock at B will not be synchronized with the clock on the ship." I'm afraid I don't understand this; since A is synchronized with B, and the ship clock is synchronized with A, then why aren't the three clocks synchronized?
This is a fairly subtle point which deserves some elaboration. A reference frame can be understood as an imaginary lattice of measuring sticks with a clock sitting at each point. Events are labeled by the position $x$ where they occur and a time $t$ as read on the clock which sits at $x$. This is crucial - the time of an event is local.
In principle, the clocks sitting at each grid point have nothing to do with each other. To make a sensible reference frame, we must synchronize them in some way, e.g. via Einstein synchronization. However, we may now imagine that a second observer moving relative to us has their own reference frame. The key insight of special relativity is that if both we and the moving observer sychronize our respective system of clocks, then each will observe the other's clocks to be out of sync.
The specific relation between our frames is given by the Lorentz transformation equations. Let my coordinates be $(x,t)$ and the moving observer's coordinates be $(x',t')$; then
$$x' = \gamma\big(x-ct\big)$$
$$t' = \gamma\big(t - \frac{vx}{c^2}\big)$$
Consider two events - the first occurs at $x=0$ and $t=0$, while the second occurs at $x=L$ and $t=0$. That is, the coordinates I assign to the first event are $(0,0$), and the coordinates I assign to the second event are $(L,0)$. The coordinates which the moving observer assigns to the first event are also $(0,0)$, but the coordinates which they assign to the second event are
$$x' = \gamma L$$
$$t' = -\gamma \frac{vL}{c^2}$$
So while I would say that the two events are simultaneous (both occurring at $t=0$), the moving observer will say that they are not (one occurs at $t'=0$, and the other at $t' = -\gamma vL/c^2$).
In keeping with your original question, let the first event at $(0,0)$ be "the ship passes the clock at $A$ when clock $A$ reads 0," while the second event is "clock $B$ reads 0." In my frame (the rest frame at clock $A$), these two events are simultaneous; because clocks $A$ and $B$ read zero at the same time, I say they are synchronized. However, according to the moving observer, the event "clock $B$ reads 0" occurs at $t'=-\gamma vL/c^2\neq 0$, which means that the clock at $B$ is not in sync with the ship's clock in their frame.
Which means the time shown on the spaceship clock is more than the time shown on the rod clock at B.
