Solution:

We claim that the required sum is $20$.

Let $T_{k}$ denote the set of all lattice points in the circle $x^{2}+y^{2}=k^{2}$. We claim that $\sum_{d \mid k}\left|S_{d}\right|=\left|T_{k}\right|$. Indeed, given a point $(x, y)$ in $T_{k}$, let $g=\operatorname{gcd}(x, y)$. Then $x / g, y / g$ are necessarily coprime, and hence $(x / g, y / g)$ visible, and $(x / g)^{2}+(y / g)^{2}=(k / g)^{2}$. This implies $(x / g, y / g) \in \bigcup_{d \mid k} S_{d}$. Next, note that the $S_{d}$'s are necessarily disjoint. Now if $(x', y')$ is a visible lattice point in $S_{d}$ where $d \mid k$, then we can write $k=g d$ so that $(x, y)=(g x', g y')$ is a lattice point in $T_{k}$. This establishes a bijection between $\bigcup_{d \mid k} S_{d}$ and $T_{k}$, and since the $S_{d}$'s are disjoint, the claim follows.

From the claim, it suffices to find the number of lattice points in the circle $x^{2}+y^{2}=(2021 \cdot 2025)^{2}$. This is equivalent to
$$x^{2}+y^{2}=3^{8} \cdot 5^{4} \cdot 43^{2} \cdot 47^{2}$$
Now it is well-known that if $x^{2}+y^{2} \equiv 0\pmod{p}$ where $p \equiv 3\pmod{4}$ is a prime, then $x \equiv y \equiv 0\pmod{p}$. Thus, we must also have $x, y \equiv 0\pmod{3^{4} \cdot 43 \cdot 47}$. It then follows that the number of lattice points is the same as the number of lattice points in $x^{2}+y^{2}=25^{2}$.

If $x=0$ or $y=0$, there are $4$ solutions. Otherwise, assume WLOG that they are both positive. Now it is well-known that all solutions to $x^{2}+y^{2}=z^{2}$ are in the form $x=g(m^{2}-n^{2})$, $y=2 g m n$, and $z=g(m^{2}+n^{2})$, where $m>n$ are coprime positive integers, and $g$ is a positive integer. Thus, we want $g(m^{2}+n^{2})=25$. Note that $g \mid 25$, so $g=1,5,25$.

If $g=25$, then $m^{2}+n^{2}=1$, so $n=0$, contradiction. If $g=5$, then $m^{2}+n^{2}=5$, which yields $m=2$ and $n=1$ and thus $g(m^{2}-n^{2})=15$ and $2 g m n=20$, so $(x, y)=(15,20),(20,15)$. If $g=1$, then we get $m^{2}+n^{2}=25$, from which we obtain $m=4$ and $n=3$. It then follows that $(x, y)=(24,7),(7,24)$, and so we have $2$ solutions when $x, y$ are both positive. This implies that there are $4 \cdot 4=16$ solutions when $x, y$ are nonzero, and so there are $4+16=20$ solutions in total.