Active transformation and passive transformation of a scalar field

Passive point of view:

Alice observes some field $\phi$ at location $x = x_0$ in her lab at Princeton USA, and finds field value $\phi(x_0)=\phi_0$. Bob observes Alice's measurement from his lab in Cambridge, UK. In his frame he sees the Princeton location as $x' = \Lambda x_0$ and confirms the same field value as Alice, which to him reads $\phi'(x') = \phi(x_0)=\phi_0$, hence $$ \phi'(\Lambda x_0) = \phi(x_0) $$

Active point of view:

Alice again observes field $\phi$ at location $x = x_0$ in her lab and finds field value $\phi(x_0)=\phi_0$. But this time Bob decides to reproduce her experiment identically in his lab and measure the same field at the exact same location relative to his frame, $x' = x_0$. All goes well and Bob finds the same value as Alice, meaning $$ \phi'(x_0) = \phi(x_0) $$ If the location of Alice's observation as seen from Bob's frame is ${\bar x}_0 = \Lambda x_0$, then conversely $x_0 = \Lambda^{-1}{\bar x_0}$ and Bob can say that $$ \phi'(x_0) = \phi(\Lambda^{-1} {\bar x_0}) $$

See for instance these notes on QFT on manifolds, particularly the following table after Eq.(8):