I know cardioid $r=1-\cos \theta$ can be parametrized to $$\gamma:[0,2\pi]\rightarrow \mathbb{R^2}, \space \gamma(\theta)=((1-\cos \theta)\cos \theta, (1-\cos \theta)\sin \theta)$$
But how is this done?
It seems like multiplying with the unit circle. But why is this "valid"?
On
As you mentioned, this is very similar to drawing a unit circle, but instead of staying constant, you're changing how far out from the origin, based on where you are around the unit circle. Does that sound familiar? It's exactly the way polar curves are constructed.
Alternatively, start with $\gamma(\theta)=(x,y)$ where x and y depend on $\theta$. This is clearly correct (literally "for a given $\theta$, plot whatever x and y correspond to that $\theta$"), but not very helpful since the formula doesn't tell us what those are.
By definition, we can substitute in $x=r \cos(\theta)$ and $y=r \cos(\theta)$ to get $\gamma(\theta)=(r \cos(\theta),r \sin(\theta))$. But we're told $r=1 - \cos(\theta)$, so we can substitute that in as well to get the parametrization given.
Also note you can see the relationships between x,y,r,$\theta$) if you draw a triangle with them, so you don't need to memorize these.
The relationship between Cartesian coordinates and Polar coordinates is $$\begin{align} x&=r\cos(\theta)&r&=\sqrt{x^2+y^2}\\ y&=r\sin(\theta)&\theta&=\arctan\mathopen{}\left(\frac{y}{x}\right)\mathclose{}+\pi\left(\frac12-\frac{x}{2\left\lvert x\right\rvert}\right) \end{align}$$
Using the left equations, and the relationship for your curve, leaves you with $$\begin{align} x&=(1-\cos(\theta))\cos(\theta)\\ y&=(1-\cos(\theta))\sin(\theta) \end{align}$$
So you have the parametrization you describe.