Introduction
Let's define buffer capacity quantitatively as
$$\beta=\cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}=-\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})}$$
that is, the relationship between concentration (in equivalents) of strong base ($c_\mathrm{b}$) or acid ($c_\mathrm{a}$) added to a solution and its change in $\mathrm{pH}$. From now on I'll assume we're adding a monoprotic base or acid, so "equivalents" and "moles" (and their concentrations) can be used interchangeably.
Pure water
Pure water has a very low buffering capacity - its $\mathrm{pH}$ is very sensitive to the addition of acids or bases. For instance, imagine we add a concentration $c_\mathrm{b}$ of base to pure water:
$\ce{H2O + B- <=> OH- + BH}$
The self-ionisation equilibrium of water will be displaced:
$\ce{[OH-]} \approx c_\mathrm{b}$
and therefore
$ c_b \approx \ce{[OH-]} = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}}$
so, taking the derivative,
$\beta_{\ce{OH-}} = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})} = \cfrac{\mathrm{d}}{\mathrm{d}(\mathrm{pH})} 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \ln{10}$
it's easy to show that, similarly, for the addition of acids,
$\beta_{\ce{H+}} = -\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})} = 10^{-\mathrm{pH}} \ln{10}$
and, combining the buffer effect of both semi-systems, we get the total buffer capacity of water:
$\beta_{\ce{H2O}} = \left( 10^{-\mathrm{pH}} + 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \right) \ln{10}$
Weak acid/base pair
1:1 solutions of weak acids/bases are a typical buffer system around $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ for the buffer. To understand why, let's imagine we add a weak acid/base pair, $\ce{HA/KA}$, with total concentration $C_\mathrm{A}$, to which we later add a certain concentration of base, $c_\mathrm{b}$. This weak acid/base pair will be described by the equilibrium
$K_A = \cfrac{\ce{[H3O+] [A-]}}{\ce{[HA]}}$
which, taking into account that $C_\mathrm{A} = \ce{[HA]} + \ce{[A-]}$, implies
$\ce{[A-]} = C_\mathrm{A} \cfrac{K_\mathrm{A}}{\ce{[H3O+]} + K_\mathrm{A}}$
From charge balance:
$\ce{[H3O+]} + \ce{[K+]} = \ce{[OH-]} + \ce{[A-]}$
Taking into account that $\ce{[K+]}$ is equal to the formal concentration of $\ce{KA}$ we added, so $c_\mathrm{b} = \ce{[K+]}$, and the previous expression for $\ce{[A-]}$:
$c_b = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} - \ce{[H3O+]} + C_\mathrm{A} \cfrac{K_A}{\ce{[H3O+]} + K_\mathrm{A}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} - 10^{-\mathrm{pH}} + C_\mathrm{A} \cfrac{10^{-pK_A}}{10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}}}$
So the buffer capacity of the solution will be:
$\beta = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}= \left(10^{\mathrm{pH}-\mathrm{p}K_w} + 10^{-\mathrm{pH}} + C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \right) \ln{10}$
Note that the first two terms are the buffer capacity of water, so the contribution of the acid/base pair is
$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \ln{10}$
Additive buffer capacity contributions can be calculated this way.
For instance, for a typical acetate buffer ($\ce{HAc} \ 0.2 \mathrm{M} $, $ \ce{NaAc} \ 0.2 \mathrm{M} $; $\mathrm{p}K_\mathrm{a} = 4.76$, $C_\mathrm{A} = 0.4\mathrm{M}$), the buffering capacity looks like this:
Ionic strength
Since buffers are moderately concentrated electrolyte solutions, the assumption that activity coefficients in them are approx. 1 shouldn't be automatic. How does this affect the calculations of buffer capacity we've performed until now?
Let's define an effective equilibrium pseudo-constant, $K'$, operating on concentrations, that is mathematically equivalent to the actual equilibrium constant, $K$, operating on activities:
$K_\mathrm{A} = \cfrac{a_{\ce{H3O+}} a_{\ce{A-}}}{a_{\ce{HA}}} \iff K'_\mathrm{A} = K_\mathrm{A} \cfrac{\gamma_{\ce{H3O+}} \gamma_{\ce{A-}}}{\gamma_{\ce{HA}}} = \cfrac{\ce{[H3O+]} \ce{[A-]}}{\ce{[HA]}}$
So reflecting the impact of ionic strength on buffer capacity becomes a question of evaluating $K'_\mathrm{A}$ and using it to replace $K_\mathrm{A}$ in our previous expressions. For instance, we can use the Debye-Hückel relationship:
$\mathrm{p}K'_\mathrm{A} = \mathrm{p}K_\mathrm{A} + A \left( 2 z_{\ce{HA}} - 1 \right) \left( \cfrac{\sqrt{I}}{1 + \sqrt{I}} - 0.1 I\right)$
where $A$ is a constant ($\approx 0.51$ at room temperature), $z_{\ce{HA}}$ is the charge of the conjugate acid and $I=\frac{1}{2}\sum c_i z^2_i$ is the ionic strength of the solution. Note that $\mathrm{p}K'_\mathrm{A}$ has to be solved recursively, however, as it depends on $I$, which depends on the concentration of the different ions, which depend on $\mathrm{p}K'_\mathrm{A}$.
A similar correction can be applied to $K_\mathrm{w}$ to obtain $K'_\mathrm{w}$.
Applying this ionic strength correction to our acetic/acetate $\pu{0.4M}$ buffer above, we can see the impact of ionic strength:
The $\ce{HCl/KCl}$ buffer
So what about the $\ce{HCl/KCl}$ buffer? Unlike the example of acetic/acetate, $\ce{HCl}$ is a strong electrolyte - so it dissociates completely at all values of $\mathrm{pH}$. So we can treat it in two ways: as simply a water buffer with a different starting $\mathrm{pH}$, or as an acid/base conjugate pair buffer with a $\mathrm{p}K_\mathrm{A} < 0$. Both are equivalent, as the acid/base contribution is negligible compared to the water contribution due to the $\mathrm{p}K_\mathrm{A}$ value.
For instance, for a typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer ($\mathrm{p}K_\mathrm{A} = -6.3$, $\mathrm{pH} \approx 0.7$, $I \approx 0.8$), the contributions to buffer capacity are as follows:
$\beta_{\ce{H2O}} = \left( 10^{\mathrm{pH}-\mathrm{p}K'_\mathrm{w}} + 10^{-\mathrm{pH}} \right) \ln{10} \approx 10^{-\mathrm{pH}} \ln{10} = 0.4594$
$\beta_{\ce{HCl/Cl-}} = C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K'_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K'_\mathrm{A}} \right)^2} \ln{10} \approx C_\mathrm{A} 10^{\mathrm{p}K'_\mathrm{A}-\mathrm{pH}} \ln{10} = 5.865 · 10^{-8}$
So, for all effects and purposes we have a buffer that behaves just like water. It is, however, a buffer that behaves just like water in a region of the $\mathrm{pH}$ scale which has a relatively large buffer capacity by virtue of being close to $\mathrm{pH}=0$. For instance, this is what buffer capacity looks like around our typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer (the red dashed line marks the initial $\mathrm{pH}$):
Note that buffer capacity is highly asymmetrical around this point - adding acid increases the buffer capacity of this system, requiring increasingly more acid to decrease $\mathrm{pH}$, while adding base decreases the buffer capacity, making the system's $\mathrm{pH}$ more sensitive to further base additions. This can be easily visualised if we realise that the area under the $\beta$ curve represents the concentration that needs to be added to move in the $\mathrm{pH}$ scale.
Note also that, unlike in weak acid/base pairs, we cannot simply increase concentration to increase buffer capacity while retaining the same $\mathrm{pH}$ - as the concentration-dependent term, corresponding to the $\ce{HCl/KCl}$ base, is negligible. Ionic strength is irrelevant for $\beta$, as well, as $K'_\mathrm{w}$ and $K'_\mathrm{A}$ is not involved - $\beta_{\ce{H3O+}}$ is the only relevant contribution to $\beta$ in this very acidic region, and, by definition, it is not affected by activity concerns (unlike $\beta_{\ce{OH-}}$ or $\beta_{\ce{HCl/Cl-}}$).
Does that mean that $\ce{HCl/KCl}$ buffers with different concentrations all have the same buffer capacity? Not at all: but their different buffer capacity is mediated by $\mathrm{pH}$. In other words, different concentrations produce different $\mathrm{pH}$ which, in turn, produces a different $\beta$ - which is unresponsive to changes in $I$ that do not change $\mathrm{pH}$.
Two final plots to illustrate this. Imagine we prepare solutions of $\ce{HCl/KCl}$ (1:1) at different total concentrations $C_{\ce{Cl-}}=c_{\ce{HCl}}+c_{\ce{KCl}}$. If we compare the two contributions to $\beta$, it's clear that $\ce{HCl/KCl}$ increases buffer capacity almost exclusively through $\mathrm{pH}$; $\beta_{\ce{H3O+}}$ is proportional to $\ce{[H3O+]}$ which increases linearly with $c_{\ce{HCl}}$ as $\ce{HCl}$ is a strong electrolyte:
Conversely, what if we control for $\mathrm{pH}$ and change only $I$? Let's imagine a series of solutions containing a fixed concentration of $\ce{HCl}$ and varying amounts of a non-reacting, strong electrolyte without common ions - such as $\ce{NaBr}$ - so we can change the ionic strength of the solution without affecting neither the $\mathrm{pH}$ nor the $\ce{HCl/Cl-}$ equilibrium. The impact of $I$ on $\beta$ would look like this (note that the minimum $I$ corresponds to a solution with $\ce{HCl}$ and no $\ce{NaBr}$):
This is because although the $\beta_{\ce{HCl/Cl-}}$ and $\beta_{\ce{OH-}}$ terms do depend on $I$, these terms are many, many orders of magnitude lower than the $\beta_{\ce{H3O+}}$ term, which doesn't.
So, after this discussion, we can now directly respond to your questions.
TL;DR
1) Why does the increased ionic strength of the solution improve water's buffer capacity?
It doesn't. The buffer capacity of solutions of strong acids are dominated by the $\beta_{\ce{H3O+}}$ term, which (by definition of $\mathrm{pH}$) is not affected by ionic strength.
There are, however, other reasons to desire an ionic strength in that range - in biological systems, many molecules, particularly proteins, have a privileged ionic strength stability gap - above or below which they tend to denaturalise or precipitate, and outside of which their biological activity can be inhibited or poisoned. Although different systems will require different ionic strengths, buffers are typically prepared with ionic strengths in the range $0.1-1\mathrm{M}$ for that reason.
2) Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases $K$?
The $\beta_{\ce{H3O+}}$ term, by construction, does not depend on $K'_\mathrm{w}$, so it is unaffected by ionic strength. The $\beta_{\ce{OH-}}$ term does, and as you mention, it is decreased at moderate ionic strength - although it is increased at higher $I$. In general, the impact of ionic strength on buffers is as follows: increasing ionic strength begins by lowering $\beta$, then, as $I$ keeps increasing, $\beta$ raises again.
3) Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl-KCl}$ is used?
Yes, any other salt could have been used to increase the ionic strength. Adding $\ce{KCl}$ affects the acid-base pair contribution to $\beta$, but as we've seen, that contribution is completely negligible - as will be the case with strong acids, which operate mainly through the $\ce{H3O+}$ contribution.
