It has to have an x and y velocity. It's just that one of them can be zero. I calculated what they were, and neither was zero.

What do you mean by the start and end of the experiment? The moment the photon leaves one mirror, and the moment it gets to the other?

I'll let t = t1 - t0, x = x1 - x0, and y = y1 - y0, where (x0, y0, t0) marks the start of the experiment and (x1, y1, t1) marks then end. You can't give just the time due to the relativity of simultaneity.

The spacetime interval is zero (it always is with light). So that's sqrt(x^2 + y^2 - t^2) = 0. Setting y to one and solving for x, you get x = sqrt(t^2 - 1). Solving for t = 2, you get x = sqrt(3). This means that, from the point of reference where the light is moving up one meter and horizontally sqrt(3), it takes two meters per c, as opposed to one in the planet's frame. The photon is moving horizontally at the same rate as the planet. Thus, if the planet moves sqrt(3) meters horizontally between the light moving from one mirror to the other, everything takes twice as long as if it doesn't move at all.