Magnetic Field Gradient of a Ring Magnet The magnetic field gradient of a ring magnet I could find online is all oriented up and down. Which means the bottom part of the ring is S or N and the top part is the other.

For simplicity I made another picture depicting the "orientation" of the dipole (from S to N).

What would the field gradient look like if I were to change the orientation of the dipole "along" the ring!? Where there is no distinct clarification between N and S!?

 A: First of all the arrangement of field lines in figure 3 is only possible around a current carrying wire or a moving charge.
Now if I understood your question well, I think you are asking about the arrangement of magnetic field lines around a magnetized ring.
There are two types of configurations these rings can have. The poles can be axially opposite or diametrically opposite.

And the field lines of the diametrically magnetized ring magnet will be like

Please note that the field lines at the contact points will always be perpendicular to the surface at that point and the field lines will always point from NORTH pole (of the magnet) to SOUTH pole outside the magnet. Inside the magnets the field lines will point in the opposite direction. Sorry I couldn't take care of these facts in my diagrams but surely, the field lines will behave like that.
A: I think at your side view, the field lines of the diametrically magnetized ring have the opposite direction inside the magnet, from S to N. And thus, at the top view, the field lines have opposite direction insede the magnet too, but not in the white field(air I suppose).
A: If you're asking what is the direction of magnetic field $\vec{B}$ for the case when dipole circles around inside the magnet. For this case $J_b=\nabla\times\vec{M}\neq0$ and is in fact same as a finite wire carrying current out of the plane of the figure. While $K_b=\vec{M}\times\vec{n}\neq0$ and it is same as a solenoid wrapped as torus. For point outside the magnet, $\vec{B}$ will same as $\vec{B}$ of finite length wire carrying current. For point inside the magnet, $\vec{B}$ can't be said to be same as $\vec{B}$ inside a toroidal solenoid.
While we're at it the dipole represented in the second figure is the correct representation of $\vec{M}$ since $J_b=\nabla\times\vec{M}=0$ while $K_b=\vec{M}\times\vec{n}\neq0$ and is equal to a straight solenoid.
Reason for a downward $\vec{B}$ at center:

we have two surfaces carrying $\vec{K_b}$ and they have their direction opposed. The larger one is ccw as seen from above while the inner one is cw.

