1. It all depends on the $\text{p}K_{\text{in}}$ of the indicator.
If we look at the Henderson-Hasselbalch equation:
$$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right) $$
You can see that when $\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and unprotonated state, because when $[\ce{A-}]=[\ce{HA}]$, the ratio between them will be 1, and $\log 1=0$.
Every time you increase $\text{pH}$ by 1, to balance the equation, the logarithm must also be 1 higher, and this happens when the thing you take the logarithm of becomes 10 times larger. This means that for a small change in pH, we have a large change in the ratio between protonated and deprotonated forms.
Typically, with indicators, you say that the colour change happens when $\text{pH} = \text{p}K_{\text{a}} \pm 1$, because outside of that interval, you will quickly have almost purely $\ce{A-}$ or $\ce{HA}$.
If you follow this far, let's try an example with an indicator with a high $\text{p}K_{\text{in}}$
You put an indicator with $\text{p}K_{\text{in}}=10$ into pure water, which has a $\text{p}K_{\text{a}}=7$. From these values, we can see that water is a stronger acid than the indicator. This means that $[\ce{H3O+}]$ just in pure water will be high enough, that the indicator is largely protonated:
$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$
$$7=10+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=-3$$
Which means $[\ce{HIn}] = 1000*[\ce{In-}]$ - try plugging in 1/1000, 1/100, 1/1, 100/1 etc. into the logarithm on your calculator to see why this is so!
Anyway, we see that when the indicator has a $\text{p}K_{\text{in}}$ significantly higher than the $\text{p}K_{\text{a}}$ of the solution, it will mostly be protonated. Why? We will see this in part 2 of the answer.
Another example. Let's see what happens if we have a low $\text{p}K_{\text{in}}$ indicator
You put an indicator with $\text{p}K_{\text{in}}=3$ into pure water ( $\text{p}K_{\text{a}}=7$). This time the indicator is a stronger acid. This means that $[\ce{H3O+}]$ is not high enough to keep the indicator largely protonated:
$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$
$$7=3+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=4$$
Which means $10000*[\ce{HIn}] = [\ce{In-}]$ - Statistically, every time you find a protonated indicator molecule, you will have found 10000 deprotonated indicator molecules!
Titration scenario: acetic acid titrated with NaOH, using an indicator with $\text{p}K_{\text{a}}=10$
Acetic acid has a $\text{p}K_{\text{a}}=4.7$. The acetic acid is a stronger acid than water, so it will have increased the concentration of $[\ce{H3O+}]$. To raise the $\text{pH}$, we need to lower this concentration. As we add $\ce{OH-}$ to the solution, the following reaction occurs:
$$\ce{H3O+ + OH- <=>> 2H2O}$$
We are removing $\ce{H3O+}$ from the solution. But the acetic acid was only partially deprotonated, and what happens if the remove $\ce{H3O+}$ from the following reaction?
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Well, by the principle of Le Chatelier, acetic acid and water reacts further, to replenish the $\ce{H3O+}$. This goes on until the acetic acid is deprotonated to a large degree, and then the $\ce{OH-}$ very quickly raises the $\text{pH}$ of the water, since there is no acid to react with anymore.
When the $\text{pH}$ gets into the area of the $\text{p}K_{\text{in}}$ of the indicator, the colour change starts to happen. It happens swiftly, because we are not using very much of the indicator. This means that it doesn't take a lot of $\ce{OH-}$ to do to the indicator what we did to the acetic acid.
2. Indicators have different equilibrium constants because they have different acid/base energetics
Indicators have different values for $pK_{in}$ for exactly the same reasons as acids have different $\text{p}K_\text{a}$. It depends on the stability of the bond to the hydrogen that can be lost as a proton. If the bond is very unstable (as in HCl), the proton will lose its share of the electrons in the bond, and become free very easily, and the compound will have a low $\text{p}K$.
For a compound such as acetic acid, protons will become free, but protons from $\ce{H3O+}$ can also transfer back to the acetic acid. The more $\ce{H3O+}$ that exists in the environment, the larger the chance, that a deprotonated acetic acid molecule will collide with it and receive the proton.
For the indicator with a high $\text{p}K_{\text{in}}$, if there is a lot of $\ce{H3O+}$, then the indicator molecules will be re-protonated as fast as they can lose their protons. Thus, they overall stay protonated.
The equilibrium constants $K_{\text{in}}$, $K_\text{a}$, etc. basically tell you how likely the molecule is to shed its proton.
