How can I deduce a lower hemisphere's boundary's orientation?

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 $0$.

You're wrong about criterion A. (Fantastic artwork, BTW.) For the lower hemisphere (think disk with downward-pointing normal) to be on your left, you must walk clockwise.