Olin is right about the problem with C23: it lowers the bandwidth of the positive feedback rendering it useless. Just decoupling the supply close to the divider will do the job.
In fact, that is probably also the reason why TI wants to keep the feedback resistance low in value. If the value is too high, the input capacitance of the op-amp alone will already lower the bandwidth of the positive feedback, so on a noisy signal the hysteresis will not work. Normally you would like the hysteresis to be faster than the response of the op-amp; this will be probably be the 510K TI talks about.
If you make sure the highest (noise) frequency that can reach the negative input will be lower than the feedback response, you should be in the clear; however, making the feedback faster then the op-amp can react will be the better option.
I guess by "complete the feedback loop" you mean "hold the inverting and noninverting inputs at the same voltage". This is basically the op-amp's only goal in life, and given suitable negative feedback, it will accomplish it. If it can't, then it will drive the output into one supply rail or the other attempting to do so.
So, why can IC1 accomplish this, while the astable multivibrator can not? Let's consider the essential components of each:
simulate this circuit – Schematic created using CircuitLab
Now consider the definition of capacitance:
$$ I(t) = C\frac{\mathrm dV(t)}{\mathrm dt} $$
It might make a little more sense algebraically re-arranged:
$$ \frac{\mathrm dV(t)}{\mathrm dt} = \frac{I(t)}{C} $$
That says, "the rate of change of current with respect to time is equal to current divided by the capacitance". So, if you put 1A through a 1F capacitor, voltage changes at a rate of 1V/s. If you increase the current or decrease the capacitance, voltage will change faster. To get voltage to change instantly, you need infinite current or zero capacitance.
For IC1, it's easy for the op-amp to respond to any change in the input. The voltage across a capacitor wants to remain constant -- it takes time and current to change it. If in some instant the input voltage increases by 1V, the output can increase by 1V, and instantly the inverting input also increases by 1V, and the two inputs have the same voltage. Mission accomplished.
But what about IC3? Say the input increases by 1V instantly. What can the opamp do? It can increase the output voltage, but the voltage across C2 (and thus, at the inverting input) can not change instantly. To change it instantly would require infinite current. But that's impossible, because the current the op-amp can drive through the capacitor is limited by R1.
So instead, the op-amp will do the best it can and saturate the output at the positive supply rail. Eventually, it will manage to charge C2 to match the voltage at the input, and the output voltage will go to 0V.
To make an astable multivibrator, you add positive feedback so as the output starts to settle to 0V the input voltage also changes. Thus, IC3 (with positive feedback added) can never accomplish its goal. It's always trying to catch up, and every time it succeeds, it starts another cycle.
Best Answer
With lots of practice, you may be able to do dynamic KVL in your head.
let's state my assumptions without any of your graphs.
Also note relevant to your previous question on same circuit, R5C1 reduces the slew rate, not the Op Amp but only after the zero crossing.
Conclusion: To make V(c1) = 50% peak of total output swing of 10.6V choose R5=R1+R3=3.5k (not 2700)