Why is the ratio thermal noise/temperature constant? I understand that more temperature means more molecules hitting each other and therefore more noise. But why is the ratio between the two roughly constant? 
If the temperature doubles, why would its Gaussian distribution also double?
 A: Consider a particle floating around in water.
The particle is constantly being hit by the water molecules moving around in thermal motion.
Approximately speaking, at each point in time, the particle gets hit left or right.
Let's simplify and say that each time the particle gets hit, it moves right one unit with probability $p$ or left by one unit with probability $q = 1 - p$.
Random walk with step size = 1
After $n$ steps, what is the probability distribution of the particle's position?
Let's think about a sequence of $n$ steps that winds up at position $k$.
In order to wind up at position $k$, it must be the case that the number of rightward steps $n_+$ plus the number of leftward steps $n_-$ is equal to $k$:
$$n_+ - n_- = k \, .$$
What's is the probability of getting a particular sequence with $n_+$ rightward and $n_-$ leftward steps?
It's $p^{n_+} q^{n_-}$, i.e. multiply by probability $p$ for each rightward step and probability $q$ for each leftward step.
Ok now how many ways can we rearrange this sequence?
There are $n$ elements, so there are $n!$ ways to rearrange it.
However, rearranging the rightward steps among themselves doesn't actually give a new sequence, so we should divide by $n_+!$, and similarly for the leftward steps.
Therefore, the probability $P(k)$ of winding up at position $k$ is
\begin{align}
P(k)
&= p^{n_+} q^{n_-} \frac{n!}{(n_+!) (n_-!)} \\
&= p^{(n+k)/2}q^{(n-k)/2} \frac{n!}{\left( \frac{n+k}{2} \right)! \left( \frac{n-k}{2} \right)!} \, .
\end{align}
This is a binomial distribution.
It has mean $\mu = n(2p - 1)$ and variance $\text{Var} = npq$.
Step size relation to temperature

If the temperature doubles, why would its Gaussian distribution also double?

Remember we started with the idea of water molecules whacking into a particle.
Let's ask: how far does the particle go each time it is hit?
First of all, the very fact that we assumed each step is independent means that we implicitly have a system with a lot of friction.
In other words, each time the particle is hit, it gains some speed, moves a bit, but then comes to rest again before the next molecular collision.
This is a good assumption for a lot of real systems that have high friction, like a particle in water.
Ok so how far does the particle go?
Well, Newton says
$$F = ma$$
and in our case, after the collision, the only force is friction $F_\text{friction} = -\gamma \dot{x}$
where $\gamma$ is the coefficient of friction.
Therefore, the equation of motion is
$$ -\gamma \dot{x} = m \ddot{x} $$
or
$$ -\gamma v = m \dot{v}$$
which has solution
$$v(t) = v(0) \exp \left( - \frac{\gamma}{m} t \right) $$
where $v(0)$ is the velocity immediately following the molecular collision.
Integrating this gives the total displacement of the particle resulting from the collision:
$$\Delta x \equiv x(t \rightarrow \infty) = \int_0^\infty v(t) \, dt = \frac{m}{\gamma}v(0) \, . $$
The initial speed $v(0)$ should be related to the kinetic energy $E$ of the molecule by something like $E \propto v(0)^2$ where the exact proportionality depends on the details of the collision geometry, etc.
Therefore, we know that the particle's displacement should be proportional to the square root of the kinetic energy of the surrounding particles.
Since kinetic energy is proportional to temperature, we have
$$\Delta x \propto T^{1/2} \, . $$
If you do a little basic math, you can show that if you scale a variable by a factor $A$, then the variance scales by $A^2$.
Therefore, if we scale our random walk step size by $T^{1/2}$ to account for the true physical size of the step, we find that the variance scales by $T$, which is what we wanted to show.
Generality of the random walk
We started with the idea of a particle getting bumped around by the surrounding water molecules, but this basic step-left/step-right process shows up all over physics.
The voltage on a resistor is the sum of the voltages from the constituent charged particles, each of which fluctuates around.
The result in that case is Johnson-Nyquest noise.
Any process that consists of independent left/right steps has the same statistical structure.
If we use some more approximations, we can show that the binomial distribution is approximately Gaussian.
To do it yourself, look up Stirling's approximation and apply it to $P(k)$.
