$$ dz_t \sim O\left(\sqrt{dt}\,\right) $$
$z$ is a Brownian motion random variable, for reference. I just don't understand what the $\sim O$ part means. I've looked up the page for Big O notation on wikipedia because I thought it might be related, but I can't see the link.
This notation is for intuition about the change in brownian motion. That is, this is not formal but is a good enough mnemonic for practitioners.
$dz_t$ is the change in brownian motion at time $t$. Lets informally think of this as $z_{t+dt} - z_t$. We know by the properties of brownian motion that this is normally distributed with variance $t+dt - t = dt$. Thus, informally again, $dz_t$ is zero-mean normal with standard deviation $\sqrt{dt}$.
That is, $dz_t$ "grows" as $\sqrt{dt}$. Loosely speaking again, for a zero mean random variable with standard deviation $c$, a lot of the probability mass is within the interval $[-c,c]$. This is captured by the Big-O notation.
In conclusion, the mnemonic is giving the intuition that if you see the $dz_t$ term (called the diffusion term) while describing a stochastic process in ito calculus, you should think of that term as being $O(\sqrt{dt})$ compared to the drift term which is O($dt$).