I am going to add a 4th point of view, hopefully complementary to the various versions based on the given that "reflection events have to be observed on the mirrors in both frames". After all, if one observer emits a light pulse in the "y-direction", the other observer must see the pulse propagating at an angle to "y" regardless of the presence or absence of a mirror setup.
The idea here has to do with two observations:
1) In any given frame light propagates not only at light speed c, but also in a direction $\vec{k}$ perpendicular to the light front at any given moment.
2) Different frames perceive different orientations of a light front due to relativity of simultaneity.
Let's see how this works. The first statement is nothing new. It follows from the Maxwell equations and it is a safe assumption. As for the second one, consider (a patch of) a plane wave light front propagating in the $y'$ direction as seen in the rest frame O' of the clock. Say it was emitted at $y'=0$ at time $ct'=0$. At later times $ct'>0$ it will be located at $y'(ct') = ct'$. Let's imagine that we validate this propagation by monitoring the light front with at least two detectors $A$ and $B$ positioned at different locations in the $y'=h$ plane, say at $(x'_A = 0, y'_A = h)$ and $(x'_B = L, y'_B = h)$, respectively. In the clock frame, both detectors will record the light front passing at $ct'= h$.
The other frame, O, must attribute the recording events at $A$ and $B$ to the passage of the same light front. But since $A$ and $B$ occur simultaneously in O', they are no longer observed simultaneously in O. Instead O sees the interactions at $A$ and $B$ at coordinates
$$
\left\{\begin{eqnarray}x_A &=& \gamma(x'_A + \beta ct'_A) = \gamma\beta h\\
y_A &=& h\\
ct_A &=& \gamma(ct'_A + \beta x'_A) = \gamma h
\end{eqnarray}\right.
$$
and
$$
\left\{\begin{eqnarray}x_B &=& \gamma(x'_B + \beta ct'_B) = \gamma(L+\beta h)\\
y_B &=& h\\
ct_B &=& \gamma(ct'_B + \beta x'_B) = \gamma(h +\beta L) = ct_A + \gamma\beta L \end{eqnarray}\right.
$$
where $\beta = \frac{v}{c}$, $\gamma = \frac{1}{\sqrt{1-\beta^2}}$, and $v$ is the relative velocity. In fact, all points of the light front that O' observes at his time $ct'$ are each observed at a different time in O, depending on their position along $x'$. Conversely, events O observes at one moment $ct$ of his time at different locations along $x$, necessarily correspond to different times $ct'$ in O'. But then, what exactly does O observe of the light front at any $ct$?
Take for instance $A$'s time in O, $ct_A = \gamma h$. We know that for $x_A = \gamma\beta h$ the light front is located at $y_A = h$. Now consider what happens at $x_B = \gamma(L+\beta h)$ at the same moment $ct_A$. The front is definitely not at $y_B = h$ because it will only arrive there at $ct_B = ct_A + \gamma\beta L > ct_A$. So it has to be at some other location $\bar{y}_B < h$. Since we know $x_B$ and $ct_A$, we can use the reverse Lorentz transformations to find the time in O' for any event $(x_B, y, ct_A)$, regardless of the exact $y$ coordinate (yes, all events occurring simultaneously in the plane $x=x_B$ at time $ct_A$ in O also occur simultaneously in O'). Once we have the corresponding $ct'$, we know the $y$ location of the front since $y' = ct'$ in O'. We obtain
$$
\bar{y}_B = ct' = \gamma(ct_A - \beta x_B) = \gamma [\gamma h - \beta\gamma(L+\beta h)] = \gamma^2 \left(\frac{h}{\gamma^2} - \beta L\right) = h - \gamma^2\beta L
$$
The two points $(x_A, y_A)$ and $(x_B, \bar{y}_B)$ now give us the orientation of the light front in O at $ct_A = \gamma h$ as
$$
\frac{y - y_A}{\bar{y}_B - y_A} = \frac{x - x_A}{x_B - x_A}\;\;\text{or}\;\;
\frac{y - h}{h - \gamma^2\beta L - h} = \frac{x - \gamma\beta h}{\gamma L + \gamma\beta h - \gamma\beta h}
$$
Simplify and we are left with
$$
y = -\gamma\beta x + \gamma^2 h
$$
In other words, as seen in O the light front is tilted in the direction of motion at a slope $-\gamma\beta$. Correspondingly, its direction of propagation in O (the normal to the front) is also tilted in the direction of motion, at a slope $\frac{1}{\gamma\beta}$.
All that is left now is to compare this direction of propagation with the one obtained from the light clock setup in the usual way. If the clock has rest length $H$, in O' a light front travels across it in time $cT'=H$. In O the corresponding travel time is time dilated to $cT = \gamma cT' = \gamma H$, during which the clock itself moves a distance $D = \beta cT = \beta \gamma H$. So the direction of the light front as seen in O is tilted in the clock's direction of motion at a slope $\frac{\Delta y = H}{\Delta x = D} = \frac{H}{\beta\gamma H} = \frac{1}{\beta\gamma}$. Done.
Note that the first derivation of the slope did not make any use of the light clock, but only of the Lorentz transformations. Detectors $A$ and $B$ are meant to single out two distinct locations on the light front at a given moment in O', but can otherwise be removed without consequences.
And I must say the wording "we affect the direction" is moot the problem is: I can't formulate it better yet.
– coobit Nov 05 '15 at 13:22but a photon... adding sideways components to a photon which was emitted perfectly up?!!! This just means to ME that emission machine affected the light speed (not in the magnitude but in direction). How is this possible?
– coobit Nov 05 '15 at 13:40