Time dilation: faster or slower? I was wondering about Time-dilation in Special Relativity. I am still a middle school student who wonders so please excuse me if I missed any important aspects. 
Let us assume we have a system of coordinates where an point $A$ and $B$ which are located upon the walls of an object where $A$ is located in the axis of $(X_1, Y_1)$ and $B$ upon axis of $(X_2, Y_1)$ are stationary and let us assume the time is the time an beam of light takes to reach from $A$ till $B$ and back.
Using the following coordinates given above we can calculate the objects length (no need for width) as being $L = X_2 - X_1$
Using the assumption we can say the ray must travel the distance of $2AB$ in order for an unit of our time to pass or since $AB$ are the line till the start and edge of object the light ray must only travel $2L$ in an stationary system of coordinates (frame of reference).
Now let us incorporate this concept of our "time unit" and apply this to a new system of coordinates where the object is travelling at an velocity $v_0$ in the $X$ axis of our system of coordinates, using the derivations of special relativity of length contraction (Lorentz-Fitzgerald Contraction) we can say the object will begin to contract in the direction of velocity ($X$ axis) with the following factor: 
$$L' = \gamma L$$
$$L' = L * (\sqrt{1 - \frac{v^2}{c^2}})$$
$$L' = (X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})$$
Using the equation of the time unit I derived for time, we can say now a unit of our time for an observer will be the distance (light will cover) for a unit of time to pass for us (observer):
$$2L'$$ 
$$2 * ((X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})) $$
This clearly shows as the object attain velocity $L'$ gets smaller super-exponentially and from Michelson-Morley experiment we know that $c$ is always constant in all frames of reference and time-frames, now using basic calculations from classical mechanics we can say that light takes less time to cover $L'$ as:
$D = \frac{L}{c}$ will always be bigger than $D' = \frac{L'}{c}$, therefore as solving this further we will get:
$$D = \frac{(2 * (X_2 - X_1))}{c}$$ 
while $D'$ gives us:
$$D' = \frac{(2* ((X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})))}{c}$$
so we can say $c$ has to cover less distance in $D'$ as $D' > D$which leads to us as a observer to calculate a quicker time on an moving body on contrary to Einsteins actual time-dilation which shows complete opposite, that shows time slows down as you increase velocity. 
Not only has Einstein been proven correct but using logical reasoning I am still confused as I still believe I should be correct yet I am not, why is my reasoning wrong? Am I missing an important factor of Special Relativity that should have been added in this thought experiment? If so what is it? and why is my line of thought incorrect? 
 A: "Using the equation of the time unit I derived for time, we can say now a unit of our time for an observer will be the distance (light will cover) for a unit of time to pass for us (observer)"
Is this observer meant to be the one who sees the object with the walls A and B moving at velocity v? If so, then although it's true this observer will see the length contracted to $ L' = L \sqrt{1 - v^2/c^2} $, it's not true that she'll see the time for the light to go from A to B and back as $2L'/c$. You're forgetting that when light is emitted from A, B is moving away from the point of emission in the observer's frame, which adds to the time on this leg of the trip, and when light is reflected from B back towards A, the A is moving towards the point of reflection, which shortens the time on this leg. 
For example, say A and B at either ends of a ruler which is 50 light-seconds long in the rest frame of A and B, and they are moving at v=0.6c in my frame. Then in my frame, the ruler is only 40 light-seconds long, so that's the distance between A and B at any given instant. But this does not mean I measure the time for light to go from A to B and back to be 80 seconds. Suppose at t=0 in my frame, A is at position x=0 and B is at position x=40, and at that moment a light flash is emitted at the position of A. Then the light will not actually catch up with B until t=100 in my frame, because after 100 seconds B has moved 100*0.6c = 60 light-seconds, and since it was at x=40 at t=0, at t=100 it will be at x=40+60=100; meanwhile of course the light moves 100 light-seconds in 100 seconds, so at t=100 it will also be at x=100. And at t=100, if B is at x=100 then A must be at x=60 at the same moment, since the distance between them is always 40 light-seconds in my frame. 
Then 25 seconds later at t=125 in my frame, A must have moved 25*0.6c = 15 light-seconds, so it will be at position x=60+15=75 light-seconds; and in those 25 seconds the light reflected at position x=100 will have moved 25 light-seconds in the -x direction, so it'll be at x=75 light-seconds as well. Thus, we conclude that in my frame it takes a time of 125 seconds for the light to travel from A to B and back to A, in spite of the fact that the distance between A and B at any given instant is only 40 light-seconds in my frame. And since the clock at A is only running at 0.8 the rate of my own clock (as measured in my frame), it will have only elapsed 125*0.8 = 100 seconds in between the light departing A and the light returning to A, which is exactly what you'd expect given that the distance from A to B is 50 light-seconds in the A/B rest frame.
If you're interested, I analyzed this example in more detail, showing how both frames manage to agree about the one-way speed of light as well as the two-way speed of light, in this answer to another question.
