So if I roll a fair six sided die twice, I can expect the probability of getting 2 6’s to be 1 in 36.
I want to start with a probability of say 1 in 100 and calculate how many times I’d need to roll a die to match that.
I get it wouldn’t be an integer (1 in 100 would be between 2 and 3), but what specifically would it be
By extension, how could I calculate the same thing but for 5’s and 6’s. My thinking in that case would be to treat it in the equation as a 3-sided die. Would that be accurate?
We're treating each die roll as independent, so to get the final probability we just multiply the probability of success at each step.
With the repeated 1-in-6 rolls: The probability of success in one step is $p = \frac 1 6$, so the probability of success $n$ times in a row is $p^n$. If you wanted to solve for $n$ to get a final probability of $\frac{1}{100}$, you could use $n = \log_p \left(\frac{1}{100}\right) = -\log_p(100)$. With $p=\frac 1 6$ that comes out to $\approx 2.57$. (Note this is similar to your guess in the comments but not exactly the same.)
If you used 2-in-6 rolls instead (success on 5 or 6) then the math would be the same but with $p = \frac 1 3$ instead of $p = \frac 1 6$.
Finally - what if you really want to get an exactly 1-in-100 result using just your 6-sided die? You can do this using a trick. First, roll the die 3 times. Let $a, b, c$ be those three results. Then define a number $X$ by $X = 36(a-1) + 6(b-1) + (c-1)$. Now $X$ is a uniformly random number from $0$ to $6^3-1 = 215$.
If $X <= 199$, we can use the last two digits of $X$ as the result from a $100$-sided die. For example, $X=157$ would count as rolling $57$ on the d100. To get the $\frac{1}{100}$ probability we wanted, we can just define success as $X=0$ (or $100$) and failure as any other value.
If $X \ge 200$, the process has failed and we need to restart. Roll 3 more times and compute a new $X$. In theory this process could go on forever (if we keep getting $X \ge 200$ but in practice we'll usually be fine even on the first set of rolls, since the probability of getting $X \ge 200$ is only $\frac{8}{216} \approx 7\%$.
You could extend this process to get any probability you want. To get $\frac{1}{N}$, just roll enough d6 so that the total number of outcomes is $\ge N$, then break the die roll outcomes into $N$ different equal groups. The leftover outcomes always mean failure and you'll need to restart the process in those cases.
Further reading if you're interested: 1, 2.