Which one of them is the time-reversed wave-function, $\psi^{\ast }\left( x,t\right) $ or $\psi^{\ast}\left( x,-t\right) $? If the wave function $\psi\left(  x,t\right)  $ is a solution of the spinless
time-independent Schr$\ddot{\mathrm{o}}$dinger  equation,
$$
i\hbar\frac{\partial}{\partial t}\psi\left(  x,t\right)  =\left[  -\frac
{\hbar^{2}}{2m}\nabla^{2}+V\left(  \mathbf{r}\right)  \right]  \psi\left(
x,t\right)
$$
then, $\psi^{\ast}\left(  x,-t\right)  $ is also the solution
$$
i\hbar\frac{\partial}{\partial t}\psi^{\ast}\left(  x,-t\right)  =\left[
-\frac{\hbar^{2}}{2m}\nabla^{2}+V\left(  \mathbf{r}\right)  \right]
\psi^{\ast}\left(  x,-t\right)
$$
and can be defined as the time reversed wave function of $\psi\left(
x,t\right)  $
$$
\psi_{r}\left(  x,t\right)  =\psi^{\ast}\left(  x,-t\right)
$$
However, in many discussions about the time-reversed operation, the time
reversed wave function $\psi_{r}\left(  x,t\right)  $ is obtained by applying
the time reversal operator $K$, which is the complex conjugate of the wave function,
$$
\psi_{r}\left(  x,t\right)  =K\psi\left(  x,t\right)  =\psi^{\ast}\left(
x,t\right)
$$
So my question is, which one is the time reversed wave function $\psi^{\ast
}\left(  x,t\right)  $ or $\psi^{\ast}\left(  x,-t\right)  ?$
The general expression for the time-reversal operator $T=UK$ (Eq. (4.4.14) in
Modern Quantum Mechanics by J. J. Sakurai), where $U$ is a unitary operator
and $K$ is the complex conjugation operator. For spinless case, one can choose
$U=1$, so $T=K$.
 A: As per your reference, it seems that you have mistaken anti-unitary operators for the time reversal operator. The time reversal operator is a kind of anti-unitary operator. The general expression for an anti-unitary operator is, as you had mentioned, on page 269 equation 4.4.14 of J.J Sakurai's book:
$$
\theta = U K
$$
Where $\theta $ is an anti-unitary operator, U is a unitary operator and K is the complex conjugation operator. 
You can't simply take U as the identity, as even though this is an anti-unitary operator, it is not necessarily the time reversal operator.
A: In the quantum mechanics the operators don't act on the functions of $(t,x)$. They act on the functions of $(x)$. So you can't define the time reversal operator as changing the direction of time. You simply define it as some anti-linear transformation that in general case may include some linear operator $\hat{T}$,
\begin{equation}
\mathcal{T}\psi(x)=\hat{T}\psi^\star(x)
\end{equation}
However this anti-linear transformation implies the reversal of the direction of time. How? The time dependence of the wavefunction in the Schrodinger picture is obtained with help of the evolution operator,
\begin{equation}
\psi_t(x)=\exp\Big(-i\hat{H}t\Big)\psi_0(x)
\end{equation}
If you act with $\mathcal{T}$ then thanks to its anti-linearity,
\begin{equation}
\mathcal{T}\psi_t(x)=\exp\Big(+i\hat{H}^Tt\Big)\mathcal{T}\psi_0(x)
\end{equation}
where $\hat{H}^T=\mathcal{T}\hat{H}\mathcal{T}$ - the time-reversed Hamiltonian. So if $\hat{H}=\hat{H}^T$ you may write $\mathcal{T}(\psi_t(x))=(\mathcal{T}\psi)_{-t}(x)$.
Similarly the evolution of the operators in the Heisenberg picture,
\begin{equation}
\mathcal{T}\hat{O}_t\mathcal{T}=\mathcal{T}\exp\Big(+i\hat{H}t\Big)\hat{O}\exp\Big(-i\hat{H}t\Big)\mathcal{T}=\exp\Big(-i\hat{H}^Tt\Big)\mathcal{T}\hat{O}\mathcal{T}\exp\Big(+i\hat{H}^Tt\Big)
\end{equation}
I.e. both this objects evolve in the reversed time direction with the time-reversed Hamiltonian $\hat{H}^T$. If you consider the operator $\mathcal{T}\hat{O}\mathcal{T}=\hat{O}$ and $\hat{H}=\hat{H}^T$ then $\mathcal{T}\hat{O}_t\mathcal{T}=\hat{O}_{-t}$.
