I'm wondering on what principles Noether's theorem is based. More precisely:
The action is a functional on the fields only. Why do we consider then variations of the space time too? In principle careful considerations, however, seem to untangle them as special field variations. So what's going on here truly?
In a comment you write
space time symmetries don't fit into the framework of the action since the action is a functional on the fields only not also on space time (space time here appears merely as a dummy variable
This isn't quite right. A given spacetime transformation often induces a transformation on fields themselves, and in this way, spacetime transformations fit into the framework of the action.
This is most easily and explicitly illustrated by way of a simple example.
Example. Consider a theory of a single real scalar field on $\mathbb R^{3,1}$ (Minkowski space). Let $\mathcal F$ denote the space of fields considered in the theory (which usually consists of e.g. smoothness assumptions and assumptions about the behavior of the fields at infinity). The action functional will be a function $S:\mathcal F\to \mathbb R$.
Now, on the one hand, the Lorentz group $\mathrm{SO}(3,1)$ acts in a natural way on $\mathbb R^{3,1}$, namely through the group action $\rho:\mathrm{SO}(3,1)\to \mathrm{Sym}(\mathbb R^{3,1})$ defined as follows: \begin{align} \rho(\Lambda)(x) = \Lambda x, \end{align} where $\mathrm{Sym}(S)$ denote the set of bijections on a set $S$. On the other hand, this group action induces an action $\rho_\mathcal F$ of $\mathrm{SO}(3,1)$ on $\mathcal F$, the space of field configurations, as follows: \begin{align} \rho_\mathcal F(\Lambda)(\phi)(x) = \phi(\Lambda^{-1} x), \end{align} which is sometimes written as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. It is this action of $\mathrm{SO}(3,1)$ on the fields that one would use to fit spacetime symmetries into the action framework. In particular, in this case we could say for example that $S$ is Lorentz-invariant provided \begin{align} S[\rho_\mathcal F(\Lambda)(\phi)] = S[\phi] \end{align} for all $\Lambda\in\mathrm{SO}(3,1)$ and for all $\phi\in\mathcal F$. All of this can also be easily extended to theories of fields of more complicated types, like vector and tensor fields. In such cases, the action $\rho_\mathcal F$ will in general be more complicated because it will contain a target space transformation.
Actually, the opposite is true: the action is never just a functional of the field components alone. It is also a functional on the coordinate differentials (and maybe also the coordinates)! Remember: $S = \int \left(x,φ,∂φ,∂^2φ,…\right) \color{red}{d^nx}$, emphasis added to make it clear. So, in general, the action can be expressed as a functional of the form $S = \int L\left(x,dx,φ,∂φ,∂^2φ,…\right)$, where the integrand $n$-form $L = d^nx$ is written with its functional dependence on $x$ and $dx$, as well as $φ$ and its derivatives made explicit.
The variational produces the following terms: $$δS = \int δL = \int \left(δx^μ K_μ + δ\left(dx^μ\right) ∧ T_μ + …\right),$$ and, in explicit terms, the variational on $d^nx$ is: $$δ\left(d^nx\right) = δ\left(dx^μ\right) ∧ ι_μ d^nx = ∂_μ\left(δx^μ\right) d^nx,$$ where $ι_μ = ι_{∂_μ}$ is the contraction operator for $∂_μ$. One of the Euler-Lagrange equations will always be: $$dT_μ + K_μ = 0,$$ where $K_μ$ is an $n$-form and $T_μ$ an $(n-1)$-form. If there's no dependence on the coordinates, themselves, then this just becomes $dT_μ = 0$ - the continuity equation for the stress tensor!
You don't normally see it, when you restrict attention to coordinate-independent field variationals, and variationals that keep the coordinates fixed, but it is part of the actual theorems posed by Noether, which covers both the case of coordinate-independent variationals (Theorem I) and the case of coordinate-dependent variationals (Theorem II); and covers primarily the case where coordinates may have non-zero variationals, though it introduces the "internal" variationals $\bar{δ}$ for coordinate-fixed variationals on field components, in passing.
But even for the case of one independent variable $\left(x^μ\right) = \left(x^0\right) = (t)$, $n = 1$ - mechanics - you still have to deal with the coordinate variational $δt$ - and by extension $δ(dt)$ - to properly handle the cases of symmetries under time translations and boosts.
