What if we throw many buckets of water into the Sun? What will happen if we throw many buckets of water into the sun and keep repeating the process? Will it get cooler or hotter?
 A: It will get hotter. More stuff will make the inside of the sun more dense from gravity alone, plus the hydrogen is fuel and will contribute to the sun burning them in the thermonuclear fusion process. And by the way, if you throw the water vertically from a distance equal to the distance to the earth, the water will accelerate to a speed of greater than 20 km/s on impact, so its kinetic energy will contribure far more to the sun already at that point than the energy required to boil that water to the temperature of the sun.
A: Conclusion
In the short term it depends on the rate at which you add the water and how you add it. The Sun could become more or less luminous as a result. In the long term the Sun will likely stay at the same temperature, will become more luminous, but not by the amount you might expect from a simple main-sequence scaling with mass, because the material you are adding is "metal-rich".
Short term effects
Let's take an example of adding 10% of the mass of the Sun in the form of liquid water (in buckets, right, but let's also assume you don't throw the buckets in). The water does not remain in the form of molecules. It is easily disassociated into atoms, then mixed in the solar convection zone and then ionised. This requires energy.
$0.1M_{\odot}$ of water contains $6.6\times 10^{54}$ molecules (about $10^{31}$ moles). With a dissociation energy of energy of 492 kJ/mol, it requires $5 \times 10^{36}$ J to produce $1.2\times 10^{55}$ H atoms and $6.6\times 10^{54}$ O atoms. These are mixed into the convection zone of the Sun, the base of which achieves temperatures of about $2\times 10^{6}$ K. To raise the atoms to this temperature requires a further $3kT/2$ per atom, so about $8 \times 10^{38}$ J.
The H atoms are comparatively easily ionised at these temperature (13.6 eV per atom) but oxygen takes a whopping 433 eV to strip 6 electrons (appropriate for temperatures of $\sim 10^{6}$ K). So to ionise the material takes a total of $5\times 10^{38}$ J.
Thus to assimilate the material into the convection zone takes around $1.3\times 10^{39}$ J, which is equivalent to the power output of the Sun for 100,000 years. Now you could argue that this energy is small compared with the "thermal reservoir" within the Sun (e.g. $1M_{\odot}$ of hydrogen at $\sim 10^{7}$ K has $5\times 10^{41}$ J, but nevertheless, the Sun's equilibrium will be disturbed if this material is added on a timescale shorter than the thermal timescale of the convection zone. This in turn is given by $\sim GM_{\odot}\times 0.02M_{\odot}/R_{\odot}L_{\odot} \sim 6\times 10^{5}$ years. If the water is added quickly, one might therefore expect the Sun to be "quenched" for a hundred thousand years as the energy residing in the outer convection zone is used up in ionising the added material.
But we also need to consider how the water is added. If it were somehow "dropped" from the orbit of the Earth it would carry angular momentum and kinetic energy. If even half the gravitational potential energy released from Earth's orbit to the solar surface were "absorbed" by the Sun, this would supply $>10^{40}$ J - easily enough to supply the heat required in the previous paragraph. In this case there would be a huge positive energy imbalance in the convection zone that would cause the Sun to expand and become more luminous for a few hundred thousand years.
Thus if the water is added quickly (less than 100,000 years!) the Sun will be put out of equilibrium - but the effects could go in either direction depending on how the water is added.
However, I will sidestep these issues and just look at what happens on longer timescales as the Sun settles down to a new equilibrium after the water has been added.
Long term effects
You might think that adding mass to the Sun would cause it to assume the main sequence configuration of a more massive star, but it is more complicated than that. If you add water, then that is mainly oxygen by mass. Partially ionised oxygen is an excellent source of opacity and would significantly change the overall "metallicity" (anything heavier than helium) of the Sun. For an extra 10% of mass added in the form of water, the metallicity would increase from being abut 1.3% metals by mass to about 9%.
As far as I know, nobody has done calculations of stellar structure for such an extreme metallicity star. In nature the most metal-rich stars are about 5 times the metallicity of the Sun (Do et al. 2018).
The effects of additional metals (particularly oxygen) are to increase the opacity of the gas and reduce the rate at which energy can be transported out of the core. The outer convection zone would also become much larger as more of the star became susceptible to convective instabilities.
The best I can do is point you to calculations for metallicities of 5% calculated by Pietrinferni et al. (2013) and then you can interpolate (or extrapolate at you own risk) appropriately. It is clear from their Fig.3, that a metal-rich $1M_{\odot}$ star is 30% less luminous and has a 10% lower surface temperature than a solar-metallicity star of the same mass. It must therefore be slightly smaller too. However, we need to do the comparison with a $1.1M_{\odot}$ metal-rich star on the main sequence. You can directly compare a metal-rich model at $1.1M_{\odot}$ with a solar-metallicity model at $1M_{\odot}$. It turns out that the star with added metals is about the same temperature, but 20% more luminous than the Sun (and therefore must be about 10% larger). I caution you though that these are scaled solar metallicity models, they do not exactly match the oxygen-rich nature of the added material here and they also include an increased He abundance to match the increased metallicity. I would also caution that I have not considered how well the metal-rich material can be mixed beneath the convection zone and into the core (the models assume the star is born from gas with that abundance).
