In general, to sample from a distribution with density $f(x,y)$ on support $\mathcal{S}$, if using a proposal distribution with density $h(x,y)$, we need to find $M$ such that
$$\sup_{(x,y) \in\mathcal{S}} \dfrac{f(x,y)}{h(x,y)} \leq M, $$
so that we can accept a proposed value with probability
$$\alpha = \dfrac{f(x,y)}{Mh(x,y)}\,. $$
Accepting with $\alpha$ is equivaluent to drawing $U \sim U[0,1]$ and accepting if $U < \alpha$.
I am assuming you understand this general premise of rejection sampling. So in this example of drawing samples from the circle using a uniform square proposal,
$$f(x,y) = \dfrac{1}{\pi} \cdot I(\underbrace{x^2 + y^2 <1}_{=\mathcal{S}}) \quad \text{ and }\quad h(x,y) = \dfrac{1}{4} I(-1 < x,y < 1)\,. $$
First, let's find $M$. In the support of $f$,
$$\sup_{x^2 + y^2 \leq 1} \dfrac{f(x,y)}{h(x,y)} = \sup_{x^2 + y^2 \leq 1} \dfrac{ I(x^2 + y^2 \leq 1)/ \pi}{1/4} = \dfrac{4}{\pi} := M\,. $$
So any proposed value from the square will be expected with probability
$$ \dfrac{f(x,y)}{M{h(x,y)}} = \dfrac{I(x^2 + y^2 \leq 1)/\pi}{M/{4}} = I(x^2 + y^2 \leq 1)\,.$$
So for any value proposed in the support of $f$, $U\sim U[0,1]$ will always be less than $1$, so we will always accept. There is thus, no need to sample from a $U$, and whenever the sampled point is inside the circle, we can accept it straightaway.