This is more a set of comments than an answer per se, but too long to fit in a comment.
Signal from oscillators: 8Vpp 1~20KHz with an offset of ~10V with a new 9V battery.
So the issue is how to couple this to another stage which can amplify it, but at the same time set an appropriate input voltage DC offset suitable to the next stage.
The obvious solution would be to use any amplifier design with reasonably high input impedance, and AC couple to it via a capacitor, so for example a cap from OSC1 to R8.
"The main problem is, on Q1 base, where the signals meet, there's no signal." Whatever voltage signal is at Q1 base will be quite small because the impedance at Q1 base will be small compared to the 1 Meg input resistors. (Especially for frequencies above the knee of the R5-C7 highpass filter.)
So the voltages at Q1 base may well be only 1/100 or 1/1000 of the signals into R8 and R9. In any case what you are more concerned with is the AC currents through R8 and R9 (and thence into Q1-base).
And probably also of concern is the DC voltage at Q1-base -- is it in a sensible range to bias Q1 to operate in it's active range, say with 3 to 4 V DC at Q1 collector? Since you have a 100k collector resistor on Q1, that suggests you are expecting a DC Ic of around 0.03mA to 0.04mA, and thus a DC voltage of rather precisely 0.03V-0.04V across R5 (and not, for example, 0.08V), but there's nothing to set a suitable voltage on Q1-base to make that happen so far as I can see.
Finally, what is the role of C9, 10nF? In parallel with R11 that appears to create a filter that will attenuate output above 160Hz or so, working to considerably suppress the signals in your range of interest, 1 kHz-20kHz.
It's difficult to say anything about what you wrote after "My mission: be able to make its output signal usable" because you don't show a schematic of what your did and it's hard to guess.
FWIW, if you feed an AC audio signal via a capacitor into a voltage follower (which has a high impedance input, hence shouldn't disrupt the source of the signal), you are going to get an output voltage that follows the input voltage. That's assuming you've set the DC level at the follower input to something reasonable. There's not much that can go wrong there, so we need to see exactly what you did that might have cause this to fail.
Bottom line, it looks like your challenge here may be simply understanding how amplifiers work (either op amps or with discrete transistors) and how to satisfy their input requirements for signal voltage or current, impedance, and DC bias (aka offset). Perhaps reading up on that topic might allow you to navigate more satisfactorily?
The first buffer outputs a clipped signal between 0.5 - 10 V losing some data by clipping it, which is shifted down to 0 - 9.5 V, still being clipped. The second buffer doesn't modify this in anyway, so to answer your question: the output would still be clipped.
Remember, voltages are only potential difference. Amplifiers reference their input signals to their ground and power supplies, so in order to prevent clipping, you could shift the first buffer's ground and voltage supply up by 0.5 V. I dont see any use for that though..
Edit:
I assume you have 10.5V < power supply available for your lever shifter for them to work. Depending what are the power supplies for the buffer amplifiers, they just might pass the 10.5 V without clipping the signals even though datasheet says otherwise.
Best Answer
Hopefully, this answer will make the reason clearer.
They are correct.
When we refer to the input impedance of a circuit, we are (almost always) referring to the small signal input impedance. That is, the input impedance gives us the ratio between how much the current will change if we make a small change in input voltage or vice versa.
$$Z_{in} = \frac{\Delta V_{in}}{\Delta I_{in}}$$
or, using the terms of calculus
$$Z_{in} = \frac{dV_{in}}{dI_{in}}$$
If we are given a circuit, such as this:
simulate this circuit – Schematic created using CircuitLab
We can calculate the relationship between the input voltage and the input current as follows.
$$I_{in} = \frac{V_{in} - V_{internal}}{R_{in}}$$
or
$$V_{in} = V_{internal} + R_{in}I_{in}$$
Differentiation gives us
$$Z_{in} = \frac{dV_{in}}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + \frac{d(R_{in}I_{in})}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + R_{in}$$
But since \$V_{internal}\$ does not depend upon \$I_{in}\$,
$$ \frac{dV_{internal}}{dI_{in}} = 0$$
So
$$Z_{in} = R_{in}$$
Which is exactly the impedance we would get if we simply treated the voltage source \$V_{internal}\$ as a short circuit.
If you understand how this works with one internal resistance and one internal voltage source, then it shouldn't be difficult to see that the same principal applies when there are a network of internal resistances and voltage sources.
The results we get "the long way" is equivalent to simply treating the internal voltage sources as shorts. Since it is also generally quicker to just do the latter, in practice, that is what we generally do. We treat internal voltage sources as shorts when calculating input impedance.