So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.)
Only about 1.5% of the sun's mass is anything other than hydrogen and helium-4. This is true all the way out from the core to the surface. We'll assume that the sun contains only hydrogen and helium-4.
All but the outermost 0.2% of the sun's mass (out to 90% of the sun's radius) is at a temperature $kT>54\,\mathrm{eV}$, which is the energy needed to turn $\mathrm{He^+}$ into $\mathrm{He^{2+}}$. (This energy is four times the Rydberg energy.) So somewhere above 99% of the sun's mass is completely ionized.
The core temperature $kT\approx 1300\,\mathrm{eV}$ is much less than the electron mass, so the matter in the core is not relativistic.
I'm going to assume that the electrons aren't degenerate; this tool (via this question) makes me think that's a pretty safe assumption for matter at the core with density $\rho \approx 150\,\mathrm{g/cm^3}$ and temperature $T \approx 10^7\,\mathrm K$.
In that case we can treat the core of the sun as a mixture of three non-interacting ideal gases, $\mathrm H^+$, $\mathrm{He}^{2+}$, and $\mathrm e^-$. As
George Herold says, each ideal gas particle has mean kinetic energy $\frac32 kT$, so we'll want the number densities. The number density for hydrogen $n_\mathrm{H}$ is
$$
n_\mathrm{H} = \rho f_\mathrm{H}/{\mu_\mathrm{H} }
$$
where $\rho$ is the mass density, $f_\mathrm{H}$ is the hydrogen mass fraction, and $\mu_\mathrm{H} = 1\,\mathrm{gram/mole}$ is the atomic mass of hydrogen. You have a similar expression for helium (with $\mu_\mathrm{He} = 4\,\mathrm{gram/mole}$). The electron number density, thanks to complete ionization, is just
$$
n_\mathrm{e} = n_\mathrm{H} + 2n_\mathrm{He}.
$$
Here's a figure showing temperature, mass density, and composition from my source above and number density as computed here:
Note that the horizontal scale (radius) is mass-weighted: you find about half the mass of the sun between 0.1 and 0.3 solar radii, so that interval takes up about half the horizontal axis. This is purely a visualization technique, so that your eye isn't distracted by the (relatively) cool, diffuse outer layers of the sun.
To find the total thermal energy density, we have to integrate. We find the thermal energy density
$$
\epsilon = (n_\mathrm{H} + n_\mathrm{He} + n_\mathrm{e})\frac32 kT
$$
and the volume of a thin shell at radius $r$ is
$$
dV = 4\pi r^2 dr
$$
This integral $\int\epsilon\, dV$ gives me a total stored kinetic energy $E=3.09\times10^{41}\,\mathrm{J}$, of which about 95% is contained within half the sun's radius.
Now, if the sun had uniform density you could estimate its gravitational potential energy, the energy that was released when all the pieces fell together, as
$$
U_\text{uniform sphere} = -\frac35 \frac{GM_\text{sphere}^2}{R_\text{sphere}} = - 2.3\times10^{41}\,\mathrm J \text{ (uniformly dense sun)}.
$$
That's pretty close to our stored heat! We can do a little bit better since we actually know the density profile of the sun, by finding the potential energy released as you lay down each spherical shell,
$$
U = - \int_0^{M_\text{sun}} \frac{G M_\text{enclosed}(r)}{r} dM = -6.15\times10^{41}\,\mathrm{J}.
$$
This gravitational self-energy is roughly twice the stored kinetic energy --- which a real astronomer would have predicted as a consequence of the virial theorem.
Why does the luminosity increase?
As core hydrogen burning proceeds, the number of mass units per particle in the core increases. i.e. 4 protons plus 4 electrons become 1 helium nucleus plus 2 electrons.
But pressure depends on both temperature and the number density of particles. If the number of mass units per particle is $\mu$, then
$$ P = \frac{\rho k_B T}{\mu m_u}, \ \ \ \ \ \ \ \ \ (1)$$
where $m_u$ is the atomic mass unit and $\rho$ is the mass density.
As hydrogen burning proceeds, $\mu$ increases from about 0.6 for the initial H/He mixture, towards 4/3 for a pure He core. Thus the pressure would fall unless $\rho T$ increases.
An increase in $\rho T$ naturally leads to an increase in the rate of nuclear fusion (which goes as something like $\rho^2 T^4$ in the Sun) and hence an increase in luminosity.
This is the crude argument used in most basic texts, but there is a better one.
The luminosity of a core burning star, whose energy output is transferred to the surface mainly via radiation (which is the case for the Sun, in which radiative transport dominates over the bulk of its mass) depends only on its mass and composition. It is easy to show, using the virial theorem for hydrostatic equilibrium and the relevant radiative transport equation (e.g. see p.105 of these lecture notes), that
$$ L \propto \frac{\mu^4}{\kappa}M^3,\ \ \ \ \ \ \ \ \ \ (2)$$
where $\kappa$ is the average opacity in the star.
Thus the luminosity of a radiative star does not depend on the energy generation mechanism at all. As $\mu$ increases (and $\kappa$ decreases because of the removal of free electrons) the luminosity must increase.
Why does the radius increase?
Explaining this is more difficult and ultimately does depend on the details of the nuclear fusion reactions. Hydrostatic equilibrium and the virial theorem tell us that the central temperature depends on mass, radius and composition as
$$T_c \propto \frac{\mu M}{R}$$
Thus for a fixed mass, as $\mu$ increases then the product $T_c R \propto \mu$ must also increase.
Using equation (2) we can see that if the nuclear generation rate and hence luminosity scales as $\rho^2 T_c^{\alpha}$, then if $\alpha$ is large, the central temperature can remain almost constant because a very small increase in $T_c$ can provide the increased luminosity. Hence if $RT_c$ increases in proportion to $\mu$ then $R$ must increase significantly. Thus massive main sequence stars, in which CNO cycle burning dominates and $\alpha>15$, experience a large change in radius during main sequence evolution. In contrast, for stars like the Sun, where H-burning via the pp-chain has $\alpha \sim 4$, the central temperature increases much more as $\mu$ and $\rho$ increase, and so the radius goes up but not by very much.
Best Answer
It is reasonably straightforward to show (e.g. see p.105 of these lecture notes) that the luminosity of a main sequence star like the Sun, depends only on its mass and its interior composition and opacity: $$ L = \frac{\mu^4}{\kappa}M^3\ ,$$ where $M$ is the mass, $\mu$ is the number of mass units per particle in the interior and $\kappa$ is the average opacity in the star.
The process of nuclear fusion turns 4 hydrogen nuclei (protons) into a helium nucleus. If we take 4 protons + 4 electrons (for neutrality), then $\mu=1/2$. For a gas of pure helium we have one helium nucleus plus 2 electrons and $\mu = 4/3$ thus as the core of the star turns from hydrogen to helium then $\mu$ increases.
Let us for the moment assume that $\kappa$ remains constant (actually the removal of electrons makes $\kappa$ decrease too, but it is not the dominant effect here), then the rate at which $\mu$ changes will be proportional to the rate of nuclear reactions, which in turn is proportional to the luminosity. Hence $$ \frac{d\mu}{dt} \propto L\ . $$ But from the first equation we can say $$\frac{dL}{dt} \propto 4\frac{\mu^3}{\kappa} M^3 \frac{d\mu}{dt} \propto 4\frac{L^2}{\mu} \propto 4 L^{7/4} \kappa^{1/4} M^{-3/4}\, , $$
Thus $dL/dt$ depends on $L^{7/4}$, which means that as the luminosity grows, the rate of change of luminosity increases significantly, since $M$ is constant.