One mistake, that I can spot, is that the relation:
$$\frac{r_{slice}}{z_{point}} = \frac{R_{cone}}{h_{cone}}$$
should be changed to
$$\frac{r_{slice}}{z_{tip}-z_{point}} = \frac{R_{cone}}{h_{cone}}.$$
I think the rest is fine.
Nice question! I learned a few things as I was looking for the solution.
I assume that the length of your light cord is $0.3\text{ meters} \times 99 = 29.7$m
(explanation: In the picture you posted, it seems that the cord starts and ends with a light bulb, so there are $99$ segments in between, each with a length of $0.3$m.
I also assume (as you state in the comments) that you want each twist to be $0.3$m apart. Note that this is not the same as a bulb being equidistant with the bulbs around it, as the bulbs on different twists will be somewhat further apart than $0.3$m. But it is a good enough approximation. Besides, I think that your original restriction (exactly equidistant bulbs) might not be possible with a conical spiral. In any case, since you are fine with each twist being $0.3$m apart, we will work with this assumption, as it makes the problem easier to solve.
The general parametric equations that define a conical spiral are:
$$\begin{array}{rl}
x =& t\cdot r\cdot \cos(\alpha \cdot t)\\
y =& t\cdot r\cdot \sin(\alpha \cdot t) \\
z =& t
\end{array}
$$
Where $t$ is a variable that expresses the vertical distance from the tip of the cone, $r$ is the radius of the cone at $t=1$ and $a$ is a parameter that affects how densely the twists are wound around the cone. The bigger the $a$ the more dense the winding.
What is $r$ in our problem? We want the height to be double the diameter so at distance $1$m from the cone tip we simply want $r = \frac14$ meters (all distance units will be expressed in meters).
What should $\alpha$ be? Setting $\alpha \cdot t = 2\pi$ means a full turn/twist around the cone, and since we want the starting point of the twist with the ending point of the twist to be $0.3$m apart, this means that $\alpha = \frac{2\pi}{0.3}$. Edit: no, this means that they are $0.3$m apart in the vertical direction ($t$ is vertical distance). What we need is that the spirals are $0.3$m apart on the surface of the cone. So, how much is $t$ if the distance on the surface of the cone is $0.3$? If we take a cross section of the cone we can form a right triangle, where the hypotenuse is $0.3$, one side (the vertical distance) is $t$, and the other side is $t/4$. Applying the pythagorean theorem we find that $t = 0.3\cdot \frac{4}{\sqrt{17}}$. So we want $\alpha \cdot \left( 0.3\cdot \frac{4}{\sqrt{17}}\right) = 2\pi \iff \alpha = \frac{2\pi}{0.3} \cdot \frac{\sqrt{17}}{4}$
We have established parameters $\alpha$ and $r$, so our conical spiral is defined fully. But how to we find the height of the cone/tree? The arc length of a conical spiral is:
$$\text{length}(t) = \frac12t \sqrt{1+r^2(1+\alpha^2t^2)}+\frac{1+r^2}{2\alpha r}\text{sinh}^{-1}\left( \frac{\alpha\cdot r\cdot t}{\sqrt{1+r^2}}\right)$$
Plugging in $\text{length}(t)=29.7$, $r=\frac14$, $\alpha = \frac{2\pi}{0.3}\cdot \frac{\sqrt{17}}{4}$ we can solve for t to get $t \approx \bbox[5px,border:2px solid red]{3.295}$ meters.
So if you make your tree about $3.3$ meters tall and make your twists about $0.3$ meters apart, then you will have the coverage that you want.
Here's how your light bulb spiral might look like (it was a bit tricky to place the $100$ bulbs on the graph, I was happy that I succeeded in the end):
$\hspace{2cm}$
And here's a side view of the spiral. As you can see there are about $11.5$ twists.
$\hspace{3cm}$
You can find the Python code I wrote to create the graphs here.
I hope this answer can help you with your lights installation. Merry Christmas! :)
Best Answer
To go from the bottom of the tree to the top $h$ increases from $0$ to $8$ feet, so $\theta$ increases from $0$ to $16\pi$. We can define $r(\theta)$ as the radius at a given $\theta$. It decreases linearly from $3$ to $0$, so $r(\theta)=3-\frac 3{16\pi}\theta$. The derivative of arc length with respect to $\theta$ can be found by Pythagoras. In $\Delta \theta$ we go up $\frac {\Delta \theta}{2 \pi}$ and we go around $r \Delta \theta=\left(3-\frac 3{16\pi}\theta\right)\Delta \theta$ Now we can integrate $$s=\int_0^{16\pi}\sqrt{\frac 1{4\pi^2}+\left(3-\frac 3{16\pi}\theta\right)^2}d\theta$$ Alpha tells me the integral is about $76.2746$. I hit the time limit before it would give me the symbolic answer. This seems reasonable as $8$ turns at the average radius of $1.5$ feet would give $75.4$ feet and the rise will add a bit.