Yes, you can still use Snell's law if u=u(z), where u in the index of refraction and z is the axis parallel to the normal of the surface.
Depending upon the length scale of the variation you may have to treat it as a differential.
The concept of "image" of a point means: the reflected/refracted light rays from it seem to come from the image point. So if you follow them backwards where they came from and prolongate them beyond the spot where they were reflected/refracted, they meet at the image point. Note: tt is not obviuos, that such a point must exist, the rays don't have to meet, and often they do only in some approximation.
The case of a mirror is a case, in which the image does definitely exist.
The rays are reflected following the rule, that the incident and outgoing angles are equal, always, independent of what media are involved!
The rays all seem to come from the reflected point.
This is a geometrical statement. That's why the concepts of "reflection" in the geometrical sense and the optical sense coinside. Those are genuinely different concepts, but the geometrical ("move the point in perpendicular direction by twice its distance") always implies the optical ("light ray from observer to reflected point defines a spot of reflexion; the angles of the rays spot-observer and spot-object are equal").
Now if you look from above the water, you see all distances in water smaller then they are (and behind the mirror there is water, too - the whole universe gets reflected). That is, the fish seems to be $\frac 5{14} H$ below the surface, the reflected fish $\frac{15}{14} H$ below the surface - and the spot of reflection sitll in the middle.
Note, that the distances seem smaler in water, not greater as you wrote. This is also a case of the concept of "image", but this time aproximate!
It's only valid for small objects when all light rays are nearly perpendicular.
Light rays are refracted to be neerer the perpendicular direction inside the wter; so the prolonged light rays seem to meet nearer the surface than the actual ones.
Best Answer
The method of apparent depth approximation is not valid here since the angles of the incident and refracted rays are not small.
To the contrary, using the Figure-01 below you have the exact solution in one stroke. Note that the result is independent of the length $\,a$.