I have a hypothetical universe where the relation between space and time is defined by this metric:
$$ds^2=-\phi^2dt^4+dx^2+dy^2+dz^2$$ Where $\phi = 1.94 \times 10^{-14}\space km\space s^{-2}$ (note that all terms are rank-2 differential form). Ignoring one of the dimensions of space, the surface of this space-time looks like this: 
I would like to describe how a photon of light travelling at c would travel on this surface, but I lack the basic skills to perform the operation using differential geometry. Note that the size of the universe is accelerating with the passage of time. I know it can be done using integration, geometry and a lot of algebra, but I'm looking for a more elegant way to solve it using the metric.
My question is, if I know the current time and the amount of time the photon has been travelling at c, can I calculate the amount of space it's traveled?
Answer
First, your metric must be corrected to be rank-2. Evidently you are trying to write a metric for the universe expanding with the acceleration $\phi$ $$\,r\sim \phi t^2$$
Your corrected metric equation would be
$$ds^2=-\phi^2t^2dt^2+dr^2$$
Where
$$dr^2\equiv dx^2+dy^2+dz^2$$
Null geodesics (trajectories of light) would be found by setting $\,ds=0$
$$r=\dfrac{\phi}{2}(t-t_o)^2$$
As you can easily see, for $\phi=0$ time in this universe stops. A better metric is
$$ds^2=-(\phi t+c)^2dt^2+dr^2$$
With the trajectories of light given by
$$r=c(t-t_o)+\dfrac{\phi}{2}(t-t_o)^2$$
Where $r(t_o)$ and $t_o$ are the coordinates of the light emission event and the negative solution is dropped as the time reversal. The coordinate distance $r$ traveled by light is
$$r=\sqrt{x^2+y^2+z^2}$$
EDIT based on OP's requirements in the comments
The features that need to be captured are
A) $x=\phi t^2$ (the diameter of the universe is proportional to the square of the time)
B) The space is closed (travel far enough in a straight line and you arrive where you started)
C) A metric that can be compared to FLRW, and hopefully one that provides a differential solution for the path of a photon
We start with a generic metric for an expanding space assuming its homogeneity and isotropy
$$ ds^2 = - c^2 dt^2 + a(t)^2 d\mathbf{\Sigma}^2 $$
Where $\mathbf{\Sigma}$ ranges over a 3-dimensional space of a uniform positive curvature and $a(t)$ is a scale factor. According to the requirement (A) where $\phi$ is acceleration
$$ r=\dfrac{\phi}{2}t^2$$
While the topology of this space may vary, the requirement (B) along with OP's comments and chart imply that the space is a 3-sphere $S^3$ whose round metric in the hyperspherical coordinates is
$$ d\mathbf{\Sigma}^2=r^2 d\mathbf{\Omega}^2 $$
Where
$$ d\mathbf{\Omega}^2=d\psi^2 + \sin^2\psi\left(d\theta^2 + \sin^2\theta\, d\varphi^2\right) $$
Combining the formulas we obtain the final metric
$$ ds^2 = - c^2 dt^2 + (\dfrac{\phi}{2}t^2)^2 d\mathbf{\Omega}^2 $$
The null geodesics equation is $ds=0$ or for $\theta=\varphi=0$
$$ \dfrac{2c}{\phi}\dfrac{dt}{t^2} = \pm d\psi $$
Solving for the path of the photon
$$ \psi=\psi_{\text{o}}\pm\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right) $$
We can set $\psi_{\text{o}}=0$ by the choice of coordinates and drop the negative direction
$$ \psi=\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right) $$
Where $t$ is the current age of the universe and $t_{\text{o}}$ is the age of the universe at the moment of the emission of the photon. Here $\psi$ represents the arc angle of the photon traveling around the expanding hypersphere of the universe. A full round trip is $\psi=2\pi$
$$ 2\pi=\dfrac{2c}{\phi}\left(\dfrac{1}{t_{\text{o}}}-\dfrac{1}{t}\right) $$
By solving for $t$ we obtain the time $t_{\text{r}}=t_{2\pi}-t_{\text{o}}$ required for the round trip
$$ t_{\text{r}}=\dfrac{t_{\text{o}}^2}{t_{\infty}-t_{\text{o}}} $$
Where $t_{\infty}$ is the age of the universe, after which a round trip at the speed of light is no longer possible, because the universe is expanding too fast
$$ t_{\infty}=\dfrac{c}{\pi\phi} $$
Finally, for the requirement (C), this metric essentially is in the same form as FLRW, just with a different scale factor, so comparing them should be pretty straightforward.
EDIT - The Hubble Law
The instantaneous $(t=const.)$ local coordinate $x$ distance along the direction of $\psi$ from a chosen location at $x_o=0$ is
$$ x=\psi r $$
Where $r$ is the current radius of the universe and $\psi=const.$ in the expansion. We can define the recession speed of the point $x$ relative to $x_o$ as
$$ v=\dfrac{dx}{dt}=\psi\dfrac{dr}{dt}=\psi\,\phi\,t=\dfrac{2x}{t} $$
Where $t$ is the current age of the universe. Thus the Hubble law is
$$ v=Hx=\dfrac{2x}{t} $$
With the Hubble parameter showing the age of the universe as double the Hubble time
$$ H=\dfrac{2}{t} $$
The radius of the observable universe (cosmic horizon) at the time $t_{\text{o}}$ is the distance, from which the photons will never reach back. From the path of the photon above for $t=\infty$ we have
$$ x=\psi r=\dfrac{2cr}{\phi t_{\text{o}}}=ct_{\text{o}} $$
Which matches the intuition. The radius of the universe when at $t_{\infty}$ it becomes larger than the observable universe
$$ r_{\infty}=\dfrac{c^2}{2\pi^2\phi}=\dfrac{\phi}{2}t_{\infty}^2 $$
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