In which direction will the ray of light travel on an $x$-$y$ plane? Let's say I am on the $x$-$y$ plane, travelling along the $x$-axis from negative towards positive. At the moment I cross the origin $(0,0)$, I throw a tennis ball in the $y$-axis at twice my speed. The resulting path of the tennis ball should look like $y=2x$ on the grid.
Let's replace the tennis ball with a photon, and I travel along the $x$-axis at the speed of $0.5c$. I emit the photon in the $y$-axis when I cross $(0,0)$. What will the path of the photon be when plotted on the grid? Is it $y=2x$?
I know that the speed of the photon is travelling away from me at the speed of $c$, and I know that the speed of that same photon is travelling away from ($0,0)$ also at the speed of $c$. But then this would mess up my graph because the distance between the photon and the origin would always be greater than the distance between it and me. Is there a way to illustrate this without messing up the graph?
 A: Let's get rid of the personal pronouns, because reference frames are important in SR and putting "you" in the experiment adds unnecessary confusion. Since it's just as easy to answer the general case as the specific case, I'll do that and then go back and answer the specific question at the end.
Max observes Sue traveling at velocity $v$ in the $+x$ direction. Sue launches a projectile, which Sue observes to move such that its position relative to her is given by $(x', y', z')$ and its velocity is the time derivative thereof, $\vec u' = (dx'/dt', dy'/dt', dz'/dt')$. What is the projectile's velocity $\vec u = (dx/dt, dy/dt, dz/dt)$ as measured by Max?
Relativistic velocities add as
$$u_x=\frac{dx}{dt}=\frac{\frac{dx'}{dt'}+v}{\left(1+\frac{v}{c^2}\frac{dx'}{dt'}\right)},\quad u_y=\frac{dy}{dt}=\frac{\frac{dy'}{dt'}+v}{\gamma_v\left(1+\frac{v}{c^2}\frac{dy'}{dt'}\right)},\quad u_z=\frac{dz}{dt}=\frac{\frac{dz'}{dt'}+v}{\gamma_v\left(1+\frac{v}{c^2}\frac{dz'}{dt'}\right)},$$
where $\gamma_v$ is the Lorentz factor
$$\gamma_v = \frac1{\sqrt{1-v^2/c^2}}$$
Derivation.
Now for the specifics:

Let $v = c/2$
Let $u' = (0, c, 0)$

Then after brief algebra:
$\vec u = (c/2, c/\gamma_v, 0)$
and
$\gamma_v = 1/\sqrt{0.75}$
Check our work by making sure the speed of light is still the speed of light:
$$u = |\vec u| = \sqrt{ (c/2)^2 + (c/\gamma_v)^2} = \sqrt{ 0.25c^2 + 0.75c^2} = c$$
So if we plotted the path of the light beam on a graph it would be given by
$y = 2\sqrt{0.75} x$

Note that for $v\ll c$, $\gamma_v \approx 1$, $v/c^2 \approx 0$, and the naieve interpretation, $\vec u = (\vec u'+ \vec v)$ is true.
A: Before we even compute anything, we can see that the trajectory is not $y=2x$ in photon case. Here's how:
Let us denote your frame of reference $S$ and the ground frame of reference $\bar{S}$. Say $S$ is moving relative to $\bar{S}$ in $x$-direction. In the $\bar{S}$ frame, events appear slower than they happen in $S$ due to time dilation. Also, the distances along $x$-direction appear shorter in $\bar{S}$ than they are in $S$, due to length contraction in $x$-direction.
Now, since there is no length contraction in $y$-direction, the distances along $y$ are the same in $S$ and $\bar{S}$. Now the effect of this on an object that is traveling in $y$-direction in $S$ is, it appears to have a smaller $y$-component of velocity in $\bar{S}$ - because the same $y$ distance appears to be traveled in a longer time in $\bar{S}$.
If you neglect this time dilation effect in $\bar{S}$ frame, which is the case with the non-relativistic tennis ball, you would have $y=2x$ in your example. But at relativistic speeds, time dilation would not be negligible, and the $y$ speed would appear smaller. So, the trajectory you would see in $\bar{S}$ frame with time dilation effect would be slightly more tilted towards the $x$-direction as compared with the non-relativistic case, i.e. $y<2x$.
So now we know that $y=2x$ is not the trajectory for photons, it has to be $y<2x$. The exact trajectory can be computed to be $y=2\sqrt{0.75}x<2x$, as mentioned in the answer by @g s.
To answer the last part of your question about illustration, you can't draw both the frames and trajectories on the same grid because we got length contraction - the grid of $S$ looks contracted in $x$-direction when seen from $\bar{S}$.
Hope this was helpful.
A: It doesn’t matter how fast you are going or what direction you are going. When you emit the photon on  the Y axis (I assume upwards) then that is the direction it will go no matter what you were doing.
