I'm leaving my original $N^{\log_23}\approx N^{1.585}$ construction below since it's easier to understand and verify, but I've found an improved $N^{\log_2(3+\sqrt{17})-1}\approx N^{1.833}$ construction:

I've increased the time per frame to $2$ seconds to make it easier to see the principle. (By the way, Mac OS Preview allows you to browse through the individual frames with the cursor keys if you download the image.)
Here's a static image of an intermediate stage:

Each branch brings forth $3$ branches of the next generation and $2$ of the generation after that, so the growth is determined by the recurrence
$a_{k+2}=3a_{k+1}+2a_k$
with characterstic values $(3\pm\sqrt{17})/2$. Thus, the number of points is $\sim k^{(3+\sqrt{17})/2}$ for $N=2k+1$, so $\underset{v,r}{\max}|S(v,r)|$ is $\Omega(N^{\log_2(3+\sqrt{17})-1})\approx \Omega(N^{1.833})$.
Here's a not quite rigorous argument that the number is in fact $\Omega(N^{2-\epsilon})$ for any $\epsilon>0$. Looking at the later stages of the constructions, you see areas that don't get filled. The reason they don't get filled is that they're close to points that are reached early and it would be more complicated to "waste enough time" to get to these areas late enough -- one would have to add paths that go back and forth a couple of times and then sprout out into these areas. But the "cost" of going back and forth a certain number of times, the number of points used up by such paths, is linear in $N$, whereas the area that can be filled using them is quadratic in $N$. So it may not be possible to efficiently fill all the small holes that are left by the final stages, but for any given pattern of paths leading into the empty areas, it can be applied at all stages except for the last $s$ stages, where $s$ depends on the "thickness" of the pattern. But these $s$ stages only give us a constant factor of $4^s$ in the number of points; we can simply drop them without changing the asymptotic dependence on $N$. So we can keep building more and more complicated paths into the empty areas, and the cost of using up a few rows of points will asymptotically not matter since the number $s$ of stages at which it becomes impossible to lay those paths is independent of $N$ whereas the number of stages at which the cost of the paths is negligible compared to the areas opened up by them grows with $N$. I believe this shows that the number is $\Omega(N^{2-\epsilon})$ for any $\epsilon>0$, though it might be a bit tedious to spell the argument out rigorously.
Here's the original $N^{\log_23}\approx N^{1.585}$ construction:

Here's a static version:

The images are being scaled down to fit the column width; you can open them in a new tab/window to get the full resolution (by right/ctrl-clicking).
All the bright points at the ends of the segments are at the same distance from the centre. Some of these are reached in different ways; I counted the number of distinct points while drawing the frames. Counting from $k=0$, and not counting the initial frame with $4$ points, there are $4(3^k+2^k)$ distinct points in the $k$-th frame, with $N=2^{k+1}+1$, so we have
$\underset{v,r}{\max}|S(v,r)|>4\cdot3^{\log_2(N-1)-1}>3^{\log_2(N-1)}\sim3^{\log_2N}=N^{\log_23}\;.$
I suspect that the number is also $O(N^{\log_23})$, but I haven't found a proof yet.
P.S.: I just realized that the construction was unnecessarily complicated -- almost the same figure can be constructed in a more straightforward way, which avoids points being reached twice and makes it obvious that the number of points is being tripled in each step:

(Again, right/ctrl-click on the image and open it in a new tab/window to get the full resultion -- it looks much nicer :-)
Here the number of points in the $k$th frame is simply $\frac433^k$.
P.P.S: I just realized that since there's automatically an edge between adjacent points, we actually need to double $N$ to be able to realize only the connections shown in the images; but that's just a constant factor so it doesn't change the $\Omega(N^{\log_23})$ conclusion.