What is the ADC Clock?
The section that you are seeing is for the clock used for the ADC. This clock is not directly related to the max sampling frequency though. The clock is what is actually being fed to the ADC module which needs to be faster than your sampling so that it can handle some magic for you.
How does the Max clock relate to the max sampling frequency?
What the datasheet is saying is that in order to get 10 bit resolution your clock can not be any faster than 200 KHz. When your clock is at that speed, you will be able to sample your signal at 15,000 samples per second.
If you don't need all 10 bits of resolution then you can provide the ADC with a faster clock and you will get a faster sampling rate, but the datasheet is not clear as to how fast you can go and still get 8 bit resolution.
I would assume that the clock to sample rate ratio is fixed, so 200K/15K = 13.33 which means you can go as low as 50 KHz clock resulting in 3.75 kSPS.
Why a minimum clock to get a 10 bit sample?
The ADC module is doing a sample and hold in which the voltage is essentially held in a capacitor. If you slow the clock down too much, the voltage can start to bleed off of the capacitor before a complete sample is performed. This change in voltage makes it such that you can't get all 10 bits accurately.
So what does this all mean?
According the the Nyquist-Shannon sampling theorem your sampling frequency needs to be at least twice the maximum frequency in your signal. You can learn more about why by looking at this question: Puzzled by Nyquist frequency
So in order to get 10 bits of resolution, the max your signal can be is 7.5 KHz, but if you need to sample a signal faster than that, you can, but the datasheet does not mention how high you can go or how much it hurts your resolution.
Circuits designed for a particular DC voltage are very unlikely to work correctly with AC, even at the same voltage. There is a very good chance the circuit will be damaged by the negative half-cycles of the AC power.
There are some exceptions to this, but they aren't really "circuits". A incandescent lightbulb works fine with AC or DC, and so does a resistive heater. Beyond that, you're heading for trouble.
Some devices that require DC power may have protection built in so that they at least don't fry with AC provided or the DC power is hooked up backwards. This is usually in the form of a "idiot diode" in series with the positive DC power lead. It simply blocks anything negative so that it can't cause harm. Another approach is a fuse followed by a clamp. If the input voltage is backwards or too high, it shorts the power leads together and blows the fuse.
Cheap consumer devices that come with their own power supply are less likely to have protection. They also then come with a warning in the instructions to use only the provided power supply. If you hook up something on you're own, they are off the hook. That's a perfectly legitimate thing for a manufacturer to do when the final sell price needs to be as low as possible.
Added:
I meant to say this earlier, but had to quit and do something else. AC of the same voltage as DC will not only have large negative voltages, but also positive voltages greater than the DC level for part of the cycle. If the AC is a sine wave, the peaks will be at ± the square root of 2 times the RMS voltage. For example, "12 VAC" sine power will go as high as +17 V and as low as -17 V at the positive and negative peaks of the cycle.
Best Answer
There is nothing like dc frequency. We call the signal which is constant and has no changes in it's value as dc signal and it has zero frequency. The bias signal is one type of dc signal. If we take Fourier representation of a sinusoidal signal like sinusoidal signal [\$\sin(\omega t)\$], it contains the odd harmonics of the fundamental frequency. there is no dc bias present in it. If this signal is applied to frequency doubler, it's frequency would be doubled and consequently the odd harmonics are also doubled.
Now consider a signal with dc bias like \$A+B\sin(2\pi ft)\$. If we consider the Fourier representation of the signal, it contains a zero frequency component of amplitude (\$A\$) and two frequency components at (\$f\$ and \$-f\$) with amplitude (\$B\$). when this signal is applied to frequency doubler it doesn't have any effect on the zero frequency but the other two frequency components get translated to \$2f\$ and \$-2f\$.
DC bias is entirely independent of the concept of frequency doubling.
Realization of signal using transforms would help to solve the frequency related questions.Hope this answered your question.