Lorentz Transformation: Message sent before finish line As she wins an interstellar race, Mavis has a “hooray” message sent from the back of her 300m long ship as she crosses the finish line at v=0.6c.  Stanley is at the finish line and at rest relative to it.  He claims the message was sent before she crossed the line.
I understand how to get the answer using the Lorentz transformation.  However, I am having trouble conceptually understanding why he observes the message before she crosses the finish line.
 A: Here's a spacetime diagram drawn on rotated graph paper (so that one can more easily measure displacements in time and space along segments and so that one can visualize the orthogonality between an observer's time and space axes).
It encodes the calculation (without explicit use of the Lorentz Transformation formulas), as well as the statements made in the earlier comments and answers. 
As @PM 2Ring says, events ev1 and ev2 are spacelike-related, and are simultaneous according to Mavis. (So, the temporal ordering of these events in frame-dependent.)
As @Ollie113 says, starting from event ev2, the light-ray wins the race to the finish line, arriving at event A, compared to Mavis's arrival at event B. (Events A and B are timelike-related. So, the temporal ordering of these events in frame-independent: B happens after A.)
In fact, no time dilation or length contraction formulas are explicitly needed.
The key relationships are that 
the area of all "light-clock diamonds" are equal and 
that the diagonals of a diamond are orthogonal... 
one diagonal represents 
an observer's tick along her time-axis and 
an observer's stick (= 1 "light-tick") along her space axis.
(For $v=(3/5)c$, the Mavis's diamond is stretched by a factor $k$ in the future-forward direction and shrunk by a factor of k in the future-backward direction, where $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=2$ is the Doppler-(Bondi k)-factor.
This construction is essentially the Lorentz Transformation in light-cone coordinates.)

One can transcribe the diamonds into Mavis' frame:

A: As I read the question, I think you already know how to calculate the answer which prompts Stanley's statement, but I will calculate it anyway because then I can answer your question which I think is primarily about the use of the term "observe". 
Let $x,t$ be coordinates in Stanley's frame, with the finish line at $x=0$. I take it that to "cross the finish line" means the front of the spaceship arrives at $x=0$.
This event is at $(0,0)$ in either reference frame. 
Now in Mavis's frame the spaceship has length 300 metres, so the event at which she sends the message is at
$(x', t') = (-300, 0)$ (assuming she is traveling in the positive direction), where the primed coordinates refer to Mavis' reference frame. In Stanley's frame this event occurs at the time
$$
t = \gamma(t' + v x'/c^2 ) = 1.25 (0 -300 \times 0.6 / c) 
\simeq -0.75\;\mu{\rm s} .
$$
So Stanley observes that the hooray message was sent before $t=0$, which means it was sent before the front of the spaceship arrived at the finishing line. To put it more fully: if we divide spacetime into 'space' and 'time' according to Stanley's system of coordinates, then the events are in this order.
Now to come to your question. I think your sense of puzzlement may be to do with a lack of clarity over the use of the term "observe". In most of physics, and certainly in special relativity, we use the word "observe" to mean "what one may deduce" and NOT "what one directly sees". Stanley receives the message after it travels to him, and he can then work out when it was sent, and he "observes" in the sense of deduces that it was emitted at time $t = -0.75\;\mu$s. 
The moment when Stanley receives the message can also be worked out. It was sent from the position
$$
x = \gamma(x' + vt') = 1.25 (-300 + 0) = -375\;{\rm m}
$$
so it takes $375/c \simeq 1.25\;\mu$s to reach the finish line, arriving there at time $-0.75 + 1.25 = 0.5\;\mu$s.
Thus Stanley's experience is that first the front of the spaceship arrives, and then a message arrives, and he can figure out the time and place from which the message was sent.
A: First note that if Mavis has received this message while she is at finishing line, it'd mean the message was sent earlier, before she cross the finishing line.
I wish you had explained it more clear than this, but do note that 
while in Mavis frame the signal had to travel 300m plus the contracted distance between the front of spaceship and finishing line, in Stanley frame however, it has to travel the contracted length of 300 (calculate it if you wish) and the distance between front of spaceship and finishing line. And, if Stanley and Mavis are not at the same location while signal is emitted, then there is an asynchronously at time of sending the signal itself. So in other word you can calculate it without lorentz transformation, it would be a little harder though. Also because we don't know the distance between the front of spaceship and finishing line when the signal is emitted, you should use the fact that she has received the message at $t=0$. I didn't use math because i am not certain whether i understand your question correctly or not. 
A: I assume that the signal is sent when she actually crosses the finish line in the ship's reference frame (the issue of how they know when to send the signal is irrelevant). 
Now, imagine that there is someone in the center of the ship that, at a time before the ship reaches the finish line, sends a signal to both, the front and the back of the ship, and that this signal reaches the ends at exactly the same time than the ships reaches the finish line. In the ship's reference frame these two events are simultaneous, they coincide with the ship reaching the finish line and with the time the message was sent.
But what does an observer stationary with the finish line sees? He sees that the light ray that moves towards the front takes a larger time to reach the front than the light that travels towards the back takes to reach the back. This is because to him, the front of the ship  moves in the same direction than the light, whereas the back of the ship moves towards the ray that moves backwards, reaching him before the front ray reaches the front of the ship. 
