This seems natural, but I can't wrap my head around it when I think about it.
When I squeeze an open bottle filled with water, the water will spill out. When I squeeze a bottle, the material collapses where I squeeze it, but expands in other areas, resulting in a constant volume. If the volume is constant, then I would think that the water shouldn't spill out.
If I were to guess, there is something related to the pressure my hand is creating inside the bottle, but I'm not entirely sure.
Squeezing the bottle does decrease its volume. Rather than a bottle, it may be more helpful to think of a full toothpaste tube; the mechanics will be the same.
If you squeeze the middle of the tube, the middle will collapse, the back will expand, and the front will expand and squirt out some toothpaste. Treating the toothpaste in the tube (or the water in the bottle) as incompressible, if the tube is full then the volume of the inside of the tube = the volume of toothpaste inside the tube. When you squeeze, some of the toothpaste comes out, indicating there is now less toothpaste inside the tube. But the tube is still full of toothpaste. Therefore, the tube volume is decreased by the amount of toothpaste that came out.
Likewise, although it may look like the collapsing middle of the water bottle is offset by the expanding top and/or bottom of the bottle, it doesn't quite do it. Assuming the bottle is completely full of water, the decrease in the bottle's volume will be easily measured by capturing the water that spills out and measuring its volume.
The surface area of the bottle is conserved, but the volume is not. Squeezing the bottle deforms it into a shape whose volume to surface area ratio is lower than it was previously.
As an example consider a bottle whose cross-section is initially a circle. The volume of the bottle will be $V_0=\pi r^2h$ where $h$ is the height of the bottle, and the surface area will be $A_0=2\pi rh$.
As we squeeze the bottle we deform it into some kind of ellipse. If we choose the major and minor axes of the ellipse to be $a=xr$ and $b=\sqrt{2-x^2}r$, then the new surface area of the bottle is given approximately by $$ A_1\approx h 2\pi\sqrt{\frac12(a^2+b^2)}=2\pi rh\quad\Rightarrow\quad\frac{A_1}{A_0}=1, $$ where we have neglected the end caps. So, the surface area of the bottle is conserved. The new volume is given by $$ V_1=\pi abh=\pi r^2h x\sqrt{2-x^2}\quad\Rightarrow\quad\frac{V_1}{V_0}=x\sqrt{2-x^2}, $$ where we have again neglected the end caps. So you can see that even though the surface area is conserved, the volume is maximum only when $x=1$. A plot of the last term, $x\sqrt{2-x^2}$ is shown below to confirm this fact.
jdj081's answer is good. I just want to address where I think you originally went wrong.
Your confusion lies in using the word "volume" in two different ways. You should differentiate between volume of the container (capacity is a better term, as jdj081 states) and volume of the liquid.
The liquid's volume doesn't change. The container's volume does. Since not all of the liquid can fit, it spills out of the container. Yes, the liquid's volume is still the same, but this does not mean the volume's container is also the same.
Just another approach to visualize it. (two dimensional) If you look on a bottle from above, you can see the round shape. A circle is the optimal shape, where you can have the biggest amount of content per package. If you start to squeeze the circle (no matter how), the amount of stuff you can put into is decreasing. Cause the amount of package is not increasing, where should it come from. (If you have an inelastic package of course.)
