There's no fixed answer to this, because whichever one you choose, if you were to reverse $u$ and $v$, you would have to choose the other one. It depends on how the $u$ and $v$ coordinates are oriented on the surface.
However, "$\color{brown}{\text{if the positive $u$ and $v$ tangent vectors at a point are oriented such that,}}$ when looking from the outside of the surface at the point, $\color{brown}{\text{the direction from $u$ to $v$ is counterclockwise, then $\partial u \times \partial v$ is the correct choice.}}$"
In order to understand orientation on this intuitive level work with the following paradigm: Assume that the $(x,y)$-plane $z=0$ of $(x,y,z)$-space is oriented "upwards", in other words: that $n=(0,0,1)$ is the "positive" normal, and consider the unit disk $D$ in the $(x,y)$-plane with its boundary $\partial D$. Then the positive orientation of $\partial D$ is for all of us the counterclockwise orientation, as seen from the tip of $n$, or from high up on the $z$-axis.
This is in accordance with your Criterion A: A man walking along $\partial D$ with his head in direction $n$, i.e., upright on the $(x,y)$-plane, has $D$ to his left.
It is also in accordance with your Criterion B: Position your right hand at $(1,0)\in\partial D$, the thumb in direction $n$. Then the fingers will curl around $\partial D$ in the counterclockwise direction, as seen from above.
Now to your lower hemisphere: A (red) man walking along $\partial L$ upright on the sphere, i.e., with the length of his body in the plane $z=0$, will have $L$ to his left when he walks clockwise, as seen by a spectator high up on the $z$-axis. The same spectator will perceive the other man walking along $\partial U$ having $U$ to his left as walking in the counterclockwise direction.
By the way: It is not necessary to cut up the sphere. It is enough to note that $\partial S^2$ for a full sphere $S^2$ is $0$.
Best Answer
If we parametrize with $x=r\cos\theta$ and $z=r\sin\theta$, then when $\theta$ increases from $0$ to $2\pi$, the graph created is a circle, but it is oriented clockwise whereas a standard (coherent) orientation would be anticlockwise, this is why we get the normal $-j$ rather than $j$.
But if we re parametrize with $x=r\sin\theta$ and $z=r\cos\theta$. When $\theta$ increases from $0$ to $2\pi$ the graph created is a circle, but this time it is oriented anticlockwise, and thus has the normal $j$.
A coherent orientation is one where, when we move around the boundary, if we were "standing" on the boundary, the enclosure must always be on our left hand side, so for a circle the coherent orientation is anticlockwise.
Imagine the orientation was clockwise, then when we "stand" on the bounday, and move around the circle clockwise, the interior of the circle is no longer on our left.
With regards to "anticlockwise", I was looking from the green direction, because this corresponds to the normal $n=-j$.