You have a few misconceptions in your reasoning. The crucial one is that the effects of a sequence of rotations are in fact not independent of each other in general, so that you can’t simply add up the effects of the individual rotations. Moreover, the order in which you apply the two rotations also matters.
The first rotation (about the $x$-axis) does indeed move the spot at which the light ray hits the plane $z=1$ from $(0,0,1)$ to $(0,1,1)$. Look a little more closely at what’s going on here: the rotation moves the original spot to some other location in the $y$-$z$ plane that’s no longer on $z=1$, and so we have to trace the ray back to this plane to find the new spot. Similarly, the second rotation (about the $y$-axis) moves the spot to some other location on a plane parallel to the $x$-$z$ plane that’s also no longer on $z=1$, so when we retrace the new ray, both the $x$- and $y$- coordinates of the intersection will have been changed by the second rotation.
As I mentioned in my comments, you really are rotating a vector here—that of the direction of the ray. The resulting intersection spot is the rotated vector scaled by the reciprocal of its $z$-coordinate so that the value of this coordinate becomes $1$. The net rotation can be represented by the product of two basic rotation matrices: $$R=\pmatrix{\cos\phi&0&\sin\phi\\0&1&0\\-\sin\phi&0&\cos\phi}\pmatrix{1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta}=\pmatrix{\cos\phi&\sin\theta\sin\phi&\cos\theta\sin\phi\\0&\cos\theta&-\sin\theta\\-\sin\phi&\sin\theta\cos\phi&\cos\theta\cos\phi}.$$ The result of rotating $(0,0,1)$ by this matrix is simply its last column, so the intersection point of the rotated ray with the plane $z=1$ is $(\tan\phi,-{\tan\theta\sec\phi},1)$. Compared to your result, there’s an extra factor of $\sec\phi$ in the $y$-coordinate. The rotations in your example correspond to $\theta=-\frac\pi4$, $\phi=\frac\pi4$, giving $(1,\sqrt2,1)$ for the shifted spot. If instead you rotate about the $y$-axis first, the spot ends up at $(\sqrt2,1,1)$.
If all you want is the resulting angle, there’s no need to project the rotated ray onto the plane. You’re rotating a unit vector that starts off pointing in the direction of the positive $z$-axis, so the cosine of the angle between the $z$-axis and the rotated vector is given by the dot product of the original vector with its image, which in this case is just the $z$-coordinate of the rotated unit vector, i.e., the lower-right entry of the rotation matrix: $\cos\theta\cos\phi$. Not entirely unsurprisingly, this value is the same regardless of the order in which you apply the two rotations.
