Having the same source for power supply and positive input for a single-supply buffer amplifier

A buffer can serve many purposes in a circuit. There are other uses besides what I mention below, but these are some fairly common ones you might encounter.

One common use is as a time delay in data transmission. It can take a bit of time for a chip to recognize the change in a signal's voltage level and react to it (such as a digital HI to LO transmission). The amount of this slight propagation delay varies from one device to the next. Sometimes it is a few nano seconds, sometimes it is much more. Also, it does take some time for the signal to propagate through the copper from one part of the circuit board to another. When there are one or multiple data signals moving from one part of the circuit to another, it may be necessary to delay one so that they arrive in a specific order or at exactly the same time.

Buffers can be used as momentary data storage. Similar to the time delay, there are situations when you need a small amount of time to store a data bit such as pipelining, but this example can be a bit difficult to explain, depending upon your knowledge of computer architecture.

Another example is signal isolation. Because a buffer is essentially a signal repeater, it can be used to isolate the signal from two parts of the circuit. This is the specific use in your example, as noted by Ignacio in his answer. In this particular case, the DAC (digital to analog converter) is sensitive to loads. A load can be anything from a resistor, speaker, IC input, or even just a piece of wire. This "sensitivity to load" means the signal may change (distort) if too much or too little of a load is seen at the output of the DAC. To prevent this, a buffer chip is used to repeat the signal from the DAC output. The output of the buffer can be connected to whatever you want (within its own limitations) without affecting or distorting the output from the DAC in any way.

Lastly, a buffer can be used a signal amplifier. This is very similar to the previous example, but the reasons for using it can be different. While the output of some chip might not be extremely sensitive to loads, it may only be able to supply a few milliamps of current. In this case, you may want to use a buffer with a larger output current rating to amplifier the original signal so it can be used to drive some larger load.

There are a bunch of separate questions in the text here, so I'll talk about those too.

My first question is that if I use both op amp of the first AD712, then will it cause any issue? Can one opamp in an IC affect the other opamp(s) in the same package?

This won't do any harm - the IC manufacturers test their parts as if you'll be using both of the components.

I understand that voltage divider with these resistances will be fairly linear. Is this correct?

Hmm. The issue I see is that you're loading your low pass filter. This isn't necessarily a bad thing, but it might not do what you expect - you no longer have an RC filter, but an R(C || R) circuit, which has a different response.

Is it possible to change your circuit so that the low pass filter is built into the buffer? For example, you could use a generic 1st order active low pass filter:

Then, your voltage divider won't have an effect on the low pass filter (since the op-amp can supply the extra current to the R||C branch, leaving your voltage divider untouched).

I also understand that I lose the resolution, as the ADC can read 0.0049 volts as-is (10bit ADC with vref = 5v) which means that the ADC should be able to read 0.01volt difference when voltage is divided (.01 at input will be 0.0040 at divider and .02 will be 0.0090 at divider).

Right - that's the point. This way your voltage steps are bigger (bad) but your maximum measurement is higher (good).

Does the above sound logical or am I missing something fundamental?

This all makes sense to me.

All Op Amps are rail to rail with supply of +12 and -12. What if I supply +5..+5.5 to last buffer op amp so that its output never exceeds its rail which can allow me to skip the schottky diodes?

This could work. Be careful, though: most rail-to-rail op-amps can only get close to their rails when they're not supplying much current to their output. If your output is +4.95 V with a +5 V rail, when your ADC pulls in a spike of current, the output will momentarily drop. I don't know how far - this depends on how much current your ADC takes and how good your op-amp is near the rail - but it's something to think about.

How much gain can I get from DC perspective without any problems?(I tried to understand the gain-vs-frequency charts but I cant get it. DC is 0 Hz or 1 Hz)

I don't really know how to answer this - it might be a stability question, and I don't have much experience there. DC is 0 Hz.

Should I put some bypass capacitors on signal line? e.g. after trans-impedance amp , put a small cap in parallel to let the high frequency signal (noise) to ground out... how do I decide the value of Capacitor such that it does not cause delay in my signal?

You're already doing this! Half of your low pass filter is a capacitor to ground.

You can choose a capacitor based on the amount of resistance around it. As you know, a resistor and capacitor in series make a low pass filter, which has some amount of phase shift associated with it. The decision is essentially a tradeoff between the cutoff frequency and how much phase shift you can tolerate at your signal's frequencies.

side note: this is an enormous question. If you can narrow down your question into something more specific, you'll get much better + faster answers. A lot of people will stop reading as soon as they see a wall of text, so the shorter your question is, the better.