I am having an hard time trying to understand why the radiated power per unit area $P$ of a black body is given by 
$$P=\frac{c}{4} u$$
in terms of the energy density $u$ and the velocity of light.

I know there is a derivation in HyperPhysics, but I did not find it particularly convincing. I cannot understand the physical significance of the parameter $\theta$ and what is its relationship with the the total energy per unit time provided by a unit area. Could someone explain it better to me?

You have successfully discovered that the kinetic energy depends on the reference frame.
That is actually true.  What is amazing, however, is that while the value of the energy is frame DEpendent, once you've chosen a reference frame, the law of conservation of energy itself is NOT reference frame-dependent -- every reference frame will observe a constant energy, even if the exact number they measure is different.  So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but you will also see that the energy before and after an elastic collision will be that same number.
So, let's derive the conservation of energy in two reference frames.  I'm going to model an elastic collision between two particles.  In the first reference frame, I am going to assume that the second particle is stationary, and we have:
$$\begin{align}
\frac{1}{2}m_{1}v_{i}^{2} + \frac{1}{2}m_{2}0^{2} &= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\\
m_{1}v_{i}^2 &= m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\end{align}$$
to save myself time and energy, I'm going to call $\frac{m_{2}}{m_{1}} = R$, and we have:
$$v_{i}^{2} = v_{1}^{2} + Rv_{2}^{2}$$
Now, what happens if we shift to a different reference frame, moving to the right with speed v?  This is essentially the same thing as subtracting $v$ from all of these terms.  We thus have:
$$\begin{align}
(v_{i}-v)^{2} + R(-v)^{2} &= (v_{1}-v)^{2} + R(v_{2}-v)^{2}\\
v_{i}^{2} -2v_{i}v + v^{2} + Rv^{2} &= v_{1}^{2} - 2 vv_{1} + v^{2} + Rv_{2}^{2}-2Rv_{2}v + Rv^{2}\\
v_{i}^{2} -2v_{i}v &= v_{1}^{2}- 2vv_{1} + Rv_{2}^{2}-2Rv_{2}v\\
v_{i}^{2} &= v_{1}^{2} + Rv_{2}^{2} + 2v(v_{i} - v_{1} - R v_{2})
\end{align}$$
So, what gives?  It looks like the first equation, except we have this extra $2v(v_{i} - v_{1} - R v_{2})$ term?  Well, remember that momentum has to be conserved too.  In our first frame, we have the conservation of momentum equation (remember that the second particle has initial velocity zero:
$$\begin{align}
m_{1}v_{i} + m_{2}(0) &= m_{1}v_{1} + m_{2}v_{2}\\
v_{i} &= v_{1} + Rv_{2}\\
v_{i} - v_{1} - Rv_{2} &=0
\end{align}$$
And there you go!  If momentum is conserved in our first frame, then apparently energy is conserved in all frames!
