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
The lithium-burning fusion reaction rate becomes significant once thermal energies are sufficient for tunneling through the Coulomb barrier between lithium nuclei and protons to form stable helium nuclei.
The first step in the hydrogen-burning (pp) chain features a lower Coulomb barrier between two protons and indeed "di-protons" are formed at lower temperatures than the threshold for lithium-burning. However, the di-proton is unstable, and to produce helium one of the protons must change into a neutron via a very slow weak force interaction before the di-proton decays. This is unlikely, and to have a big enough population of di-protons transforming into deuterons via this weak force interaction requires a much higher temperature (by an order of magnitude) than just the threshold temperature for proton fusion.
In fact a more relevant comparison is the temperature thresholds for lithium-burning versus deuterium-burning. The latter takes place at $\sim 10^6$ K, a factor of three lower than for lithium-burning and exactly what one would expect given the three-times smaller Coulomb barrier.