How a diode-detector affects to Frequency Response?
But what does detector in the frequency domain?

Something really complicated.  We get trained to think about signal processing almost exclusively in the frequency domain, and we forget that the frequency domain stuff was only invented as a mathematical trick to make it easier to understand what's going on in the time domain.  If it's easier to do in the time domain, do it there.
In fact, envelope detection is really horrid from a frequency domain point of view, because not only are there nonlinearities in play, but there's nonlinearities and that perfectly linear capacitor that are interacting.  So generally if you want to analyze AM radio demodulation you try to find some equivalent operation that is linear, or at least less horridly nonlinear.
In this case, just considering the demodulation to be clipping isn't bad.  Figure that the clipping operation can be denoted as $y(t) = f\left( x(t) \right )$, where $f(x)$ is (thankfully) memoryless.  So you can assert* that it has a Taylor's expansion $$f(x) = \sum_{n=0}^\infty h_n x^n$$ where the $h_n$ are the terms of the expansion**
If we feed a sum of sinusoids through $f(x)$, the above expansion tells us that we're going to get just about every possible distortion product out of the thing.  We'll get stuff at the sum of all the sinusoid frequencies, at every possible combination of sum and difference frequencies, and at ever possible combination of the sum and difference of every possible harmonic of both sinusoids.
At this point, I'm just going to claim without proof that the element at the difference between the carrier and the sideband frequencies will be prevalent -- if the carrier and the modulating signal are widely spaced apart.  AM radio works because you're transmitting at 500kHz or more, and modulating the carrier with a signal that only occupies 0Hz to 3000Hz or so.
I'm not going to try to prove this in more detail because the math gets crazy in the frequency domain -- and because you can so clearly understand the operation in the time domain.

Why the example in audio editor doesn't work?

First, because you just multiplied $\cos \omega_c t$ with $\cos \omega_s t$  For AM modulation, you want something like $\left(\cos \omega_c t\right ) \frac{s(t) + 1}{2}$ -- you multiply the carrier by something that never falls below zero.
Second, because your carrier and your modulating tone are so closely spaced (although the signal may be recognizable).
Here's some plots.

Top, a picture of one 1kHz cycle's worth of your signal, as you've defined it.  Because you didn't shift your modulating tone, you don't have enough carrier to form an envelope.

My version with your frequencies, your modulating tone shifted by 1 and reduced by half.

My version with a carrier frequency of 20kHz, and the modulating tone shifted.



* You can assert this, but you'd be lying, because it has derivatives that go to infinity -- but you can make up something arbitrarily close to a clipping function that does have a Taylor's expansion -- just assume I've done that.
** And I use $h$ in honor of the horrid things I'm doing to the math...