This drawing I made when I was thinking about the problem shows that my initial idea was simply to calculate the nearest point to the circle whats center point is closest to the target, then from that point (as a new target), calculate the nearest point to the other circle. (The bright green lines shows this).
problem visualized quite crudely
In this video, in the first half you can see it's working as intended, but in the second half, when the intersection reduces to theoretically a single point, the errors are apparent.
This other video shows (likely) another problem near the poles. It looks very much like gimbal-lock but I'm using quaternions and slerp to interpolate these positions, so it shouldn't be. Also, without the limiting factor it places the object to the right place. (But it's not using interpolation then, because it's not needed)
Your idea (as far as I understand it from the picture and description) is not correct. For example, by your description and picture, for point 6 your algorithm will find the nearest point as a point that is on the circumference of the bigger circle but not the nearby intersection of the circles, which it should be.
Some nomenclature: $c_1$ is one considered circle, with $M_1, r_1$ being its center and radius. Similiarly, $c_2, M_2, r_2$ for the second circle, center, radius. $A$ and $B$ are the points where $c_1$ and $c_2$ intersect. $P$ is the point whose nearest intersection point we are looking for. Spherical distance between points $X,Y$ is $d(X,Y)$.
My solution would go as follows:
1) Determine if $P$ is inside both circles $c_1$ and $c_2$. If yes, $P$ is inside the intersection and is itself the nearest point, and we are done.
2a) Find the nearest point $N_1$ to $P$ on the whole $c_1$. As commented on by Jean Marie, use the great circle through $M_1$ and $P$ to do that. If $P=M_1$, choose any point $N_1$ on $c_1$.
2b) Do the same as 2a) for circle $c_2$ and find $N_2$.
3a) Determine if $N_1$ is on the arc of $c_1$ that borders your intersection. If yes set $d_1:=d(N_1,P)$, otherwise set $d_1$ equal to $+\infty$.
3b) Do the same as 3a) for $N_2$ and $c_2$, defining the distance $d_2$.
4) Find $d_{min}=\min (d_1, d_2, d(P,A), d(P,B))$
5) Your nearest point is whichever corresponds to the argument that produced the minimum in 4).