Magnetic Boundary Conditions in case of surface current

I'm sorry, but I'm not familiar with the equation you've given. In this instance, I'd use this equation: $$ mathbf { H^{||}_{above} - H^{||}_{below} = K times hat n } $$

With $ mathbf { H^{||}_{below} = 0 , H^{||}_{above} } $ becomes the cross product of the surface current and the normal to the surface. As you've mentioned, it has a magnitude of 2 A/m. The direction is $ mathbf { hat x times hat y } $ which gives $ mathbf { hat z } $.

I hope this helps!

Edit: Fixed a superscript error.

In your equation, $vec{n_{12}}$ is pointing upward (+ve y). So, $vec{H_1}$ should be pointing outside of the screen (which is +ve z) so that $vec{H_1} times vec{a_{12}}=vec{K}$ is pointing in the +ve x-direction.

An easier way to see this is to consider a loop enclosing the current. And using right-hand rule, the field must point outside of the screen in the air region.