Why do objects traveling faster appear shortened rather than elongated? I'm sure this seems like a stupid question but I hope someone can try to explain it as if to a child...
This relates to Lorentz contraction. Why do objects in motion relative to the observer not get longer?
In the train example. A stationary observer is at the middle of a train car which is traveling from the observer's left to his right. A flashbulb goes off in the middle of the train car as it passes the observer.
Now, I understand the speed of light is constant from the observer's POV regardless of the speed of the source. So the observer sees the light hit the left side of the car before it hits the right. 
If the car is going 0.99x the speed of light, the observer at rest sees the light reach the left side of the car in (let's say?) 1/100th the time it would take if the car was at rest relative to the observer -- so since we're measuring the distance by the constant speed of light, the left end is shorter. That is essentially what the Lorentz contraction is, isn't it? So my question is: If that's true, then why wouldn't the observer also see it take 99x longer for the light to catch up with the right end of the car -- making the car appear exponentially longer until, if it were going at the speed of light, it would appear infinitely long to the right of the observer and infinitely short to his left?
In trying to understand this, I found these two diagrams on Wikipedia, and what I noticed was that the bottoms of the vertical sticks aren't in the same place. And if they were, then more speed would make the left side shorter approaching zero, and the right side longer approaching infinity.


 A: Let's say that at the outset, the observer on the platform has a wristwatch reading 12:00.  He considers his own location to be point $0$.  He considers the right side of the train car to be at location $x$.  
The light flashes.  The train car continues to move.  The light catches up to the right side of the train car at (say) 12:01 by the observer's wristwatch, at location $y$ (farther to the right than $x$).
You are saying that if the train car is going very fast, $y$ is going to be very large.  That is true.   But $y$ has little to do with how long the observer thinks the train car is.  
The observer says "At 12:00, the light flash went off at point $0$.  At 12:00, the right end of the train was at $x$ (not $y$)!!.  Therefore, at 12:00, the distance from the light flash to the right side of the train car was $x$."  
This is intended to explain why your calculation is wrong, which is what you asked about.  It does not explain how to get it right, but there are many many many standard sources for that.
A: First of all, the train is moving at constant speed (inertial frame of reference) so, to observers within the train, it must behave as if it were stationary.  This means that the light takes the SAME amount of time to reach the left and right edges of the train.
This can be objectively verified by putting two synchronized clocks, one at each end of the train, such that the clocks freeze when light hits them.  Both clocks must have identical readings at the end of the experiment, and this objective reality (clock reading) must be explained by all observers inside and outside the train.
From this identical clock reading, and the constancy of speed of light, we conclude that the path of the light beam must have the same length on the left and right halves for all observers.
Imagine a train which is twice as long as high, the light source is at the top of the midpoint, and the clocks are at the two opposite bottom corners.  Imagine an external observer standing at the midpoint of the train.  If the train were stationary, the light beam would have a 45 degree angle on both sides.
But, since the left clock is moving rightwards, the light beam will have a smaller angle, say 30 degrees from vertical.  Given basic geometry, the only way for the beam in the right half to have the same length is for it to have an angle less than 45 degrees from the vertical also.  This can only happen if the leading (right) edge is lagging behind where it "should" be, i.e. the front of the train is shortening.
A: Actually your perception is correct in regards to the events you consider, but the conclusion does not contradict the concept of length contraction. 

In short, what you are looking at are simultaneous events in the train's rest frame. These do measure the length of the train in the rest frame, but not its contracted length as seen by the 'stationary' observer. The reason is that with respect to the stationary observer these events are no longer simultaneous  due to relativity of simultaneity. In fact, as you correctly observe, to the stationary observer they appear separated by an interval of time that goes to infinity as the speed of the train approaches that of light. Correspondingly, the distance between them as seen by the stationary observer also goes to infinity, and there is nothing wrong with that! 
So where does length contraction come from? Actually it comes from the fact that the stationary observer measures the train length, as he observes it, by using events that are simultaneous in his stationary frame, but not simultaneous in the train's frame.   

At this point I prefer to back up my statements with a bit of math to avoid misunderstandings, since there seem to be plenty when it comes to this topic. 
Let O be the stationary observer and O' be the train's observer (at rest in the train's frame) in relative motion at velocity $v = \beta c$ ($\beta = \frac{v}{c}$, $\gamma = \frac{1}{\sqrt{1-\beta^2}}$). Take the origin of O' in the center of the train as usual and let A be the left (rear) end of the train, and B the right (forward) end. Let the length of the train be $2L$, such that the positions of A and B wrt O' are $x_A' = -L$ and respectively $x_B'=L$. 


*

*In his frame, O' measures the train's length by using events that are simultaneous at A and B. We could argue that this is not necessary since A and B are at rest wrt O', but the procedure must be the same regardless of whether it is applied to objects at rest or in motion. Since there is no better way to properly define it for objects in motion, we also extend the procedure as such for objects at rest. So lets say that O' measures the distance between A and B using events $\mathcal{A} = (-L, ct'_0)$ and $\mathcal{B} = (L, ct'_0)$ at his time $ct'_0$. What does the stationary observer O see instead? To him event $\mathcal{A}$ has coordinates 
$$ 
x_A = \gamma(x_A' + \beta ct'_0) = \gamma(-L + \beta ct'_0)\\
ct_A = \gamma(ct'_0 +\beta x_A') = \gamma(ct'_0 -\beta L)
$$ 
while event $\mathcal{B}$ occurs at 
$$
x_B = \gamma(L + \beta ct'_0) = x_A + 2 \gamma L\\ 
ct_B = \gamma(ct'_0 + \beta L) = ct_A + 2 \beta \gamma L
$$ 
Obviously he observes $\mathcal{A}$ and $\mathcal{B}$ occurring at a distance 
$$
x_B - x_A = 2 \gamma L \equiv \gamma( x_B' - x_A') > ( x_B' - x_A') = 2L
$$ 
and separated in time by 
$$
c\Delta t = ct_B - ct_A = 2 \beta \gamma L \equiv \beta (x_B - x_A) > 0
$$
These are the two events you were considering and, again obviously, they don't show a length contraction, but a length dilation! Yep, but this length dilation occurs between non-simultaneous events, so it cannot measure the train length as it is observed by O.

*Let's see how O really measures the train's length. For this purpose O has to consider events or observations $\bar{\mathcal{A}} = (\bar{x}_A, ct_0)$ and $\bar{\mathcal{B}} =  (\bar{x}_B, ct_0)$ that are simultaneous with respect to him at some time $ct_0$, although in O' they still occur at the same locations $x_A' = -L$ and $x_B' = L$, at the train's ends. This means that the relation between the coordinates of $\bar{\mathcal{A}}$ and $\bar{\mathcal{B}}$ in the two frames must be
$$
\bar{x}_A = \gamma ( -L + \beta c\bar{t}_A')\\
ct_0 = \gamma (c\bar{t}_A' -\beta L)
$$
and
$$    
\bar{x}_B = \gamma ( L + \beta c\bar{t}_B')\\
ct_0 = \gamma (c\bar{t}_B' + \beta L)
$$
If we eliminate the O' times $c\bar{t}_A'$ and $c\bar{t}_B'$, and then look at the distance between $\bar{\mathcal{A}}$ and $\bar{\mathcal{B}}$ as observed in O we obtain
$$
\bar{x}_B - \bar{x}_A = \gamma(L + \frac{\beta}{\gamma}ct_0 - \beta^2 L - (-L) - \frac{\beta}{\gamma}ct_0 - \beta^2 L) = \gamma (1- \beta^2) 2L 
$$ 
or 
$$
\bar{x}_B - \bar{x}_A = \frac{2L}{\gamma}
$$

In other words, O does see the train length contracted by a factor of $\frac{1}{\gamma}$, although he sees distances between events that are simultaneous in O' as dilated by a factor of $\gamma$.

