The Schrödinger equation is not necessarily symmetric under time reversal. For a time reversal operator $\mathcal{T}$, we have $$\mathrm{i}\hbar\frac{\partial}{\partial t}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t)$$ only when $$\left[H,\mathcal{T}\right]=0.$$
Consider an observable $A$. We have $$\mathcal{T}A\mathcal{T}^{-1}=A'.$$ We require that the following relations hold:
a) Correspondence of position: $$\mathcal{T}x\mathcal{T}^{-1}=x^\prime=x$$
b) Correspondence of momentum (the sign change is because in a classical sense, the particle would be moving in the opposite direction under time reversal): $$\mathcal{T}p\mathcal{T}^{-1}=p^\prime=-p$$
c) Commutation relation: $$[x_i,p_j]=\mathrm{i}\hbar\delta_{i,j}\Rightarrow\mathcal{T}[x_i,p_j]\mathcal{T}^{−1}=[x_i^\prime,p_j^\prime]=[x_i,−p_j ]=−[x_i,p_j]$$
So, $$\mathcal{T}(\mathrm{i}\hbar\delta_{i,j})\mathcal{T}^{-1}=-\mathrm{i}\hbar\delta_{i,j}\Rightarrow\mathcal{T}\mathrm{i}\mathcal{T}^{-1}=-\mathrm{i}.$$
This suggests that we 'construct' the time reversal operator in the following manner, using some linear unitary operator $U$ and an antiunitary operator $\mathcal{K}$ which performs complex conjugation here: $$\mathcal{T}=U\mathcal{K};\quad\mathcal{K}\mathrm{i}\mathcal{K}^{-1}=-\mathrm{i}.$$ In other words, the time-reversal operation in quantum mechanics involves both complex conjugation and negating the time: $$t\rightarrow -t;\quad \mathrm{i}\rightarrow-\mathrm{i}.$$
Let $t^\prime=-t$, such that $\mathcal{T}\psi(x,t)=\psi^\prime(x,t^\prime)$. Note that this is not the same definition as the quoted section in the question; we do not take $\psi(x,t^\prime)=\psi^\prime(x,t^\prime)$ (I suppose that there may be additional commentary regarding that equation beneath the sections included in the quotation). Requiring that $\psi^\prime(x,t^\prime)$ is a solution to the corresponding Schrödinger equation, \begin{align}\mathrm{i}\hbar\frac{\partial}{\partial t^\prime}\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime)&\Leftrightarrow\mathrm{i}\hbar\frac{\partial}{\partial t^\prime}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t)\\\mathcal{T}\mathrm{i}\hbar\mathcal{T}^{-1}\mathcal{T}\frac{\partial}{\partial t}\mathcal{T}^{-1}\mathcal{T}\psi(x,t)&=\mathcal{T}H\mathcal{T}^{-1}\mathcal{T}\psi(x,t)\\U\mathcal{K}(\mathrm{i}\hbar)U^{-1}\mathcal{K}^{-1}U\mathcal{K}\left(\frac{\partial}{\partial t}\right)U^{-1}\mathcal{K}^{-1}\psi^\prime(x,t^\prime)&=\mathcal{T}H^\prime\mathcal{T}^{-1}\psi^\prime(x,t^\prime)\\&=H^\prime\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime)\end{align}Thus,$$\boxed{-\mathrm{i}\hbar\frac{\partial}{\partial t}\psi^\prime(x,t^\prime)=H\psi^\prime(x,t^\prime);\quad-\mathrm{i}\hbar\frac{\partial}{\partial t}\mathcal{T}\psi(x,t)=H\mathcal{T}\psi(x,t).}\tag{A}$$Equation (A) holds irrespective of $[H,\mathcal{T}]$.
