The physicist’s answer would be correct anywhere, no matter the temperature or density of the fluid.

Anywhere with gravity sufficient to overcome the surface tension of the water from forming around the object or clinging to the container.

Nope, the rise in water shows exclusively the volume. Density is derived using volume and mass, with mass being measurable by weighing. The only factor density plays directly is indeed floating, but you can just push the sphere under with a thin rod without significantly impacting the measurement.

It's generally simple - water has some volume, sphere has some volume. Assuming a cylindrical container, the base of the cylinder is known, and the volume of it is equal to the volume of water, which means you can derive the height (or vice versa). If you now add the sphere, the sphere pushes out the water, which means you add the volume of the sphere to the water - and if you measure the height, you can calculate the total volume of water+sphere, and subtract the initial volume of water to isolate the volume of the sphere.

If you fully submerge an object, you displace its VOLUME of water. It will always be the same amount (imagine a bowling ball and a soccer ball). You might need to push it.

When you relate that to its weight, you get the density.

Imagine you have two 1kg lumps of steel. Out of one you make a bowl with thin walls the out of the other a metal ball. Afterwards they still have (about) the same weight, volume and density - as it's still steel.

But put the bowl onto the basin and with a little luck it will float like a boat *, put in the steel ball it will sink.

But when you put in the bowl sideways without splashing it will displace about the same amount of water than the ball. And it will still weigh 1kg.

*= The thing with floating people remember from school: The floating bowl will displace as much water as it weighs - that's the thing Archimedes found out. So our 1kg bowl will displace 1kg of water - which is incidentally about 1 liter, because the density of water is about 1kg/l (it was intentionally defined that way, some guys in the French revolution used water to define mass by volume)

attach something to the bottom, pull, fix position

Edit: I am late. kudos for everyone who was faster :)

Density comes into it if we are measuring buoyancy. The weight of the displaced water is equal to the weight of the buoyant object, and the ratio of volume submerged vs above water should give a way to determine density, but I don't have the precise method in my head right now.

However assuming all you want is the volume, you simply need to fully submerge the object (by applying enough downward force if its density is too low), and the displaced volume is equal to the objects volume. The displaced water will have to flow to the top of the container, so by measuring the rise in water level and multiplying with the base area of your container you can get it easily.

Just make sure you don't accidentally also measure your hand if you're pushing down the sphere.

Submerging the sphere gives you the volume. You're thinking of the legend of Archimedes who measured volume this way and measured mass with an ordinary scale and deduced the density from those two values.