In my opinion, it's better to reason with a **position-vs-time graph** (a spacetime diagram). 

To start you off, below is one using the usual convention where **t runs along the horizontal axis** <BR>*(although many relativity discussions use time t along the vertical axis)*.

https://www.desmos.com/calculator/c2mtuzt6ty
[![robphy-Desmos-c2mtuzt6ty][1]][1]

I was following you until... 

> Therefore, it takes two seconds for the light beam to 'pull ahead' of the vehicle by 300,000km.

In addition, it's not clear to me why you say

> However, when I plug 0.5c into the time dilation formula, I find that the observer in the spacecraft does not measure one second to pass for every two seconds that pass for the observer outside.

In any case, I think it is best to draw a diagram to describe your scenario.

By the way, the arithmetic for calculations in relativity is easier if one chooses a relative velocities of $\frac{3}{5}c$ or $\frac{4}{5}c$, which lead to right triangles with pythagorean triples. (This does not happen for $\frac{1}{2}c$ or $0.99c$.)

The mathematics of time-dilation is related to trigonometry with adjacent side of a right-triangle. This fact and a diagram often help guide my physical intuition.


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By the way, one way to get the 8.660254 as the elapsed time along the spaceship worldline is by
$$\sqrt{(15)(5)}=10\left(\frac{\sqrt{3}}{2}\right)=8.660254...,$$
where $5$ and $15$ are "radar-times" from a radar measurement made by the lab. <BR>
Note: $\Delta t=(15+5)/2=10$ and $\Delta x/c=(15-5)/2=5$ <BR>
and thus $(\Delta t+\Delta x/c)=15$ and $(\Delta t-\Delta x/c)=5$<BR>
so that $(15)(5)=(\Delta t+\Delta x/c)(\Delta t-\Delta x/c)=
(\Delta t^2-(\Delta x/c)^2)=(10)^2-(5)^2
$.
 

  [1]: https://i.sstatic.net/S5Bnl.png