Figure below shows a shampoo bottle in a perfect cylindrical shape...

Mass of bottle $= {m}_{0}$

Length of bottle $= L$

base Area $= A = \pi{r}^{2}$

density of shampoo $= ρ$

mass of shampoo $= ρfAL$

Center of mass of system

$y = \dfrac{{m}_{0}\dfrac{L}{2} + \left(ρfAL\right)\left(\dfrac{fL}{2}\right)}{{m}_{0} + ρfAL}$

for critical angular displacement, $mg$ will pass through tilted side.

From the diagram $tan\theta = \dfrac{r}{y}$

$tan\theta = \dfrac{r\left({m}_{0} + ρALf\right)}{\dfrac{L}{2}\left({m}_{0} + ρAL{f}^{2}\right)}$

at $f = 0 \text{ and } f = 1$, tipping angle $\theta$ will be same.

for very small values of $f$, we can neglect ${f}^{2}$ terms

$⇒ tan\theta = \dfrac{r}{\dfrac{L}{2}}\dfrac{\left({m}_{0} + ρALf\right)}{{m}_{0}}$

$\theta = \tan^{- 1}\left(\dfrac{\dfrac{r}{L}}{2}\dfrac{\left({m}_{0} + ρALf\right)}{{m}_{0}}\right)$

So if $f$ increases $\theta$ will increase