Complex conjugate of the Schrodinger Equation

Question

My question is from the mathematical part of Quantum Mechanics, so I will try my best to create enough context.
This is the one-dimensional time-dependent Schrodinger Equation: $$i\hbar \frac{\partial \psi}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \psi }{\partial x^2} +\displaystyle {\hat {V}\psi}$$
$i,\hbar,m$ are certain physical constants and $\psi$ is a function of $x$ and $t$.
$\displaystyle {\hat {V}}$ on the other hand is the potential energy operator.
My textbook takes the complex conjugate of this equation(note that $\psi^*$ is the conjugate of $\psi$), $$-i\hbar \frac{\partial \psi ^*}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \psi ^* }{\partial x^2} +\displaystyle {\hat {V}\psi^*}$$
and multiplies it by $\psi$ $$-i\hbar\psi \frac{\partial \psi ^*}{\partial t}= -\frac{\hbar^2\psi}{2m}\frac{\partial^2 \psi ^* }{\partial x^2} +\displaystyle {\hat {V}|\psi|^2}$$
My Question: How can $\psi \displaystyle {\hat {V}\psi^*}= \displaystyle {\hat {V}|\psi|^2}?$Does it not depend on the nature of $\displaystyle {\hat {V}}?$
For example, if $\displaystyle {\hat {V}}$ was $-i\hbar\frac{\partial}{\partial x}$, given that $\psi = e^{i(kx-\omega t)}$($k$ and $\omega$ are constants),
$\psi \displaystyle {\hat {V}\psi^*}=-e^{i(kx-\omega t)} i \hbar(ik e^{-i(kx-\omega t)})=\hbar k$ whereas
$\displaystyle {\hat {V}|\psi|^2} = \displaystyle {\hat {V}1}=0!$
Or is $\displaystyle {\hat {V}|\psi|^2}$ just a notation?