My question follows chapter 10 of Tung's Group Theory book, in particular definitions 10.11 and 10.12. Let $\mathcal{R}[\Lambda]$ be an $n\times n$ matrix representation of $L_+^{\uparrow}$ and a wave-function $\{\Psi^a\}$, which transforms (actively) under Lorentz as
$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\equiv\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x).\tag{1}$$
I think I understand this: under active Lorentz transformation $x\mapsto\Lambda^{-1}x:$, $\mathcal{R}[\Lambda]\Psi(x)\overset{\text{passive}}{\equiv}{\Psi}'(\Lambda x)$ becomes $\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x)\equiv{\Psi}'(x)$, where the equivalence is due to passive transformations involving changes of frames, as you can see from the image below
On the other hand, when we consider operator-fields Tung writes
$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\overset{\text{op. tr.}}{:=}U[\Lambda]\Psi(x)U[\Lambda^{-1}]\equiv\mathcal{R}[\Lambda^{-1}]\Psi(\Lambda x).\label{eq2}\tag{2}$$
In think the last expression should be
$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\overset{\text{op. tr.}}{:=}U[\Lambda]\Psi(x)U[\Lambda^{-1}]\equiv\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x).\tag{3}$$
I know I am missing something, because \eqref{eq2} is similar to 5.1.6-7 in Weinberg's book QFT vol. 1, but involving also translations.