This is very simple and I can 50% understand it but would like to properly understand why it is.
If we have an infinitesimal Lorentz transformation
$\Lambda^\mu _\nu = \delta^\mu _\nu + \omega^\mu _\nu$
(sorry, the indices shouldn't be directly under one another on some terms.)
with $\omega^\mu _\nu$ infinitesimal.
Here's the part I can't completely comprehend. The transformation on a scalar field is given by
$\phi(x) \rightarrow \phi'(x) = \phi(\Lambda^{-1}x)=\phi(x^{\mu}-\omega^\mu _\nu x^{\nu})=\phi(x^{\mu})-\omega^\mu _\nu x^{\nu}\partial_{\mu}\phi(x)$
Firstly, what is the inverse of $\Lambda$ in terms of $\delta$ and $\omega$? And how how has that final term arisen? Is it because $\omega$ is so small $\phi(\omega^\mu _\nu x^{\nu}) \approx \omega^\mu _\nu x^{\nu}$? But why the differential after that then?
Inverse is such that multiplication of a Lorentz transform and its inverse gives identity, then you work it out by hand (to first order only). It should be the same but with the minus sign $\Lambda^\mu _\nu = \delta^\mu _\nu - \omega^\mu _\nu$.
The final term comes from the Taylor expansion as omega is infinitesimal, so you keep only first term. Note this is multivariate expansion as we have a vector. It could help if you think about it like $\omega^\mu _\nu x^{\nu} = a^\nu$ where $a^\nu$ is small. Then you can expand this scalar field by this small value around x.
$\Lambda(t)^{\mu\;}_{\;\;\nu}=\Lambda(0)^{\mu\;}_{\;\;\nu}+t\Lambda'(0)^{\mu\;}_{\;\;\nu}+\mathcal{O}(\Lambda^2)=\delta^{\mu}_{\;\;\nu}+t\;\omega^{\;\mu}_{\quad\nu}+\mathcal{O}(t^2)$
$(\Lambda^{-1})^{\mu}_{\;\;\nu}=\eta^{\mu\alpha}\eta^{}_{\nu\beta}\Lambda^{\beta}_{\;\;\alpha}$
$\phi({\bf x}-t{\bf a})=\phi({\bf x})-t{\bf a}\cdot \nabla \phi\;\;\;\;\ $ c.f. $\;\;\; \phi(x+y)=\phi(x)+y\frac{d\phi}{dx}$
So you eventually get $\phi(x^{\mu}-\omega^{\mu}_{\;\;\nu}x^{\nu})=\phi(x^{\mu})-a^{\beta}\partial^{}_{\beta}\phi(x^{\mu})$. Hope this helps. It may not be entirely clear how Taylor works here by analogy...You could look up some more general Taylor formulations but it is not worth it, I assume this is QFT, I think that's what you mostly do there.