In this question I will be considering what is called in physics a "classical Lagrangian field theory" from a geometric point of view.
One is given an $n$ dimensional (smooth, real) "spacetime manifold" $M$, with some "field" $\psi$ on it. From a purely local point of view, the field is assumed to have components $\psi^i(x)$. A (first-order) Lagrangian is then an $n$-form $$ \hat L=\mathcal L(\{x^\mu\},\{\psi^i(x)\},\{\partial_\mu\psi^i(x)\})dx^1\wedge...\wedge dx^n $$ that depends on the field, the field derivatives and maybe on the spacetime coordinates explicitly.
In physics, it is considered pathological for a Lagrangian to depend explicitly on the spacetime coordinates. Usually it signals a "nonclosed" system, where there are externally imposed fields, whose dynamics are not controlled by a Lagrangian.
From a more precise point of view, let $E\rightarrow M$ be a rank $k$ vector bundle over $M$ (it can be a general fiber bundle, and in fact, it is often not a vector bundle, but even in those cases, one can often consider it a vector bundle - for example the bundle of metrics, which is nonlinear may be enlarged to the bundle of symmetric (0,2) tensors, which is).
The field $\psi$ is considered a section of $E$. In a local trivialization $E$ has coordinates $(\{x^\mu\},\{y^i\})$.
We can consider the first jet bundle $J^1E$ of (sections of) $E$, whose local coordinates are $(\{x^\mu\},\{y^i\},\{y^i_\mu\})$.
If $\psi$ is a section of $E$, it "looks like" in local coordinates as $$\psi: x\mapsto (\{x^\mu\},\{\psi^i(x)\}), $$ and its first jet prolongation $j^1\psi$ looks like as $$ j^1\psi:x\mapsto (\{x^\mu\},\{\psi^i(x)\},\{\partial_\mu\psi^i(x)\}).$$
The (first-order) Lagrangian is then a horizontal $n$-form on $J^1E$, which in a local trivialization looks like as $$ \hat L=\mathcal L(\{x^\mu\},\{y^i\},\{y^i_\mu\})dx^1\wedge...\wedge dx^n $$ where the $dx^\mu$ are considered 1-forms on $J^1E$.
The fields are not involved here, however, the action functional is defined as $$ S[\psi]=\int (j^1\psi)^\ast\hat L, $$ where $$(j^1\psi)^*\hat L=\mathcal L(\{x^\mu\},\{\psi^i(x)\},\{\partial_\mu\psi^i(x)\})dx^1\wedge...\wedge dx^n.$$
The problem:
It is a well known fact that if $E\rightarrow M$ is some fiber bundle (like the $E$ previously defined, or $J^1E$), then if $f:E\rightarrow\mathbb R$ is some smooth function (the target space of this map is not really relevant), and in some local trivialization this function looks like as $$ f(\{y^i\}), $$ eg. $f$ doesn't depend on the base coordinates $x^\mu$, then this is not a trivialization-independent statement. In some other trivialization, it will look like as $$ f^\prime(\{x^{\prime\mu}\},\{y^{\prime i}\}). $$
On the other hand, as I have stated in the beginning of the question, in physics it is considered pathological to have a Lagrangian that depends explicitly on $x^\mu$. However by the previous argument, it should not be possible to construct a Lagrangian on $J^1E$, that looks like $$ \hat L=\mathcal L(\{y^i\},\{y^i_\mu\}) $$ in all trivializations.
Now, I know that in physics, there is often a lot of background structure assumed. These background structures can break "general covariance". I can illustrate this with the following example: For the massless Klein-Gordon field in Minkowski-spacetime, the Lagrangian (in physics language - so this is already pulled back via the field's prolongation) is $$ \mathcal L=\frac{1}{2}\eta^{\mu\nu}\partial_\mu\psi\partial_\nu\psi\ dx^1\wedge...dx^n, $$ where $\eta^{\mu\nu}$ is the canonical "flat" form of the dual Lorentz-metric. It is not explicitly $x$-dependent, but this is not a trivialization-independent statement. If we perform a non-Poincaré coordinate transformation, the Lagrangian will look like $$ \mathcal L=\frac{1}{2}g^{\mu\nu}(x)\partial_\mu\psi\partial_\nu\psi\sqrt{|\det g(x)|}dx^1\wedge...dx^n. $$
Here the general form of the metric $g_{\mu\nu}(x)$ is $x$-dependent, but is an externally imposed field, so the Lagrangian is $x$-dependent.
However, in physics, if we account for everything, then we can get rid of all background structures. For example, we can consider the gravitational field (the metric) $g$, a matter field $\psi$ (a Dirac field, or a scalar multiplet), and a gauge connection $A$, and mesh everything into a glorious Lagrangian $$ \mathcal L(\psi,\partial\psi,A,\partial A,g,\partial g,\partial\partial g) $$ (this Lagrangian will be second order in $g$, but that doesn't change the problem).
Now, this Lagrangian will never acquire explicit $x$-dependence, through either diffeomorphisms of $M$, or changes of trivialization in $E$ (here $\psi$ is a section of $E$, and $E$ is considered a $G$-bundle for some Lie group $G$, and $A$ is a $G$-connection) or for that matter, in the "connection bundle" in which $A$ lives.
So from this, it seems to me that if we use fiber products to create one massive fiber bundle $\mathcal E$ from all these bundles involved, then the unified Lagrangian is still a horizontal element in $\Omega^n(J^1\mathcal E)$, yet it will never acquire explicit $x$-dependence.
Question:
Considering all the lengthy exposition I have given above, how do you invariantly and geometrically characterize Lagrangians which do not depend explicitly on the base manifold's coordinates? Since this should be a trivialization-dependent statement, yet it also seems to be that it is possible to construct Lagrangians that absolutely do not depend on any background structures, and are as such $x$-independent.