What does it mean for an action to be invariant under $x \to x'$, $\phi \to \phi'$?

1. Noether's theorem works even for non-geometric theories, so to be as general and simple as possible, we shall not use notions & concepts from differential geometry. For the purpose of Noether's theorem, it is enough to discuss infinitesimal variations: $$ \delta x^{\mu} ~:=~ x^{\prime\mu} - x^{\mu} ~=~ \varepsilon~ X^{\mu}(x),\tag{1}$$ $$ \delta\phi^{\alpha}(x) ~:=~\phi^{\prime\alpha}(x^{\prime})-\phi^{\alpha}(x) ~=~ \varepsilon~ Y^{\alpha}(\phi(x),\partial\phi(x), x),\tag{2}$$ where $\varepsilon$ is an infinitesimal ($x$-independent) parameter, and $X^{\mu}$ and $Y^{\alpha}$ are generators.

2. If $V~\subseteq~\mathbb{R}^4$ is a spacetime region, let $$ V^{\prime}~:=~\{ x^{\prime}\in \mathbb{R}^4 \mid x \in V \} ~\subseteq~\mathbb{R}^4 \tag{3}$$ denote the varied spacetime region.

3. The infinitesimal variation of the action is by definition $$\delta S_V~:=~ S_{V^{\prime}}[\phi^{\prime}] -S_V[\phi]~:= ~ \int_{V^{\prime}}\! d^4x^{\prime}~{\cal L}(\phi^{\prime}(x^{\prime}),\partial^{\prime}\phi^{\prime}(x^{\prime}),x^{\prime})-\int_V\! d^4x~{\cal L}(\phi(x),\partial\phi(x),x).\tag{4}$$ Formula (4) is $S^1_{111}-S^0_{000}$ in OP's notation. See e.g. Refs. 1 & 2.

4. The infinitesimal variation (1) & (2) are called a quasi-symmetry of the action if the infinitesimal variation (4) is a boundary integral, cf. my Phys.SE answer here. In the affirmative case, Noether's theorem leads to an on-shell conservation law.

References:

1. H. Goldstein, Classical Mechanics, 2nd edition, Section 12.7.

2. H. Goldstein, Classical Mechanics, 3rd edition, Sections 13.7.