Replacing 2022 with $2n$, the answer is $2n^2 - 2n + 2$.

Construction The construction for $n=5$ shown on the left half of the figure easily generalizes for all $n$.

First proof of bound Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$.

Claim — The distance of the leg of the journey $A_{2i-1} \to A_{2i} \to A_{2i+1}$ is at most $2n - d_i$.
Proof. Of the two arcs from $A_{2i-1}$ to $A_{2i+1}$, Bunbun will travel either $d_i$ or $2n-d_i$. One of those arcs contains $A_{2i}$ along the way. So we get a bound of $\max(d_i, 2n-d_i) = 2n-d_i$. $\square$

That means the total distance is at most
$$\sum_{i=1}^{n} (2n - d_i) = 2n^2 - (d_1 + d_2 + \dots + d_n).$$

Claim — We have
$$d_1 + d_2 + \dots + d_n \ge 2n - 2.$$
Proof. The left-hand side is the sum of the walk $A_1 \to A_3 \to \dots \to A_{2n-1} \to A_1$. Among the $n$ points here, two of them must have distance at least $n-1$ apart; the other $d_i$'s contribute at least 1 each. So the bound is $(n-1) + (n-1) \cdot 1 = 2n-2$. $\square$

Second proof of bound Draw the $n$ diameters through the $2n$ arc midpoints, as shown on the right half of the figure for $n=5$ in red.

Claim (Interpretation of distances) — The distance between any two points equals the number of diameters crossed to travel between the points.
Proof. Clear. $\square$

With this in mind, call a diameter critical if it is crossed by all $2n$ arcs.

Claim — At most one diameter is critical.
Proof. Suppose there were two critical diameters; these divide the circle into four arcs. Then all $2n$ arcs cross both diameters, and so travel between opposite arcs. But this means that points in two of the four arcs are never accessed — contradiction. $\square$

Claim — Every diameter is crossed an even number of times.
Proof. Clear: the diameter needs to be crossed an even number of times for the loop to return to its origin. $\square$
This immediately implies that the maximum possible total distance is achieved when one diameter is crossed all $2n$ times, and every other diameter is crossed $2n-2$ times, for a total distance of at most
$$n \cdot (2n - 2) + 2 = 2n^2 - 2n + 2.$$