A closed $k$-form is called intrinsically harmonic if there is some Riemannian metric with respect to which it is harmonic. E. Calabi (Calabi, Eugenio, An intrinsic characterization of harmonic one-forms, Global Analysis, Papers in Honor of K. Kodaira 101-117 (1969). ZBL0194.24701.)
showed that a one-form having non-degenerate zeros on a compact manifold without boundary is intrinsically harmonic if and only if it satisfies a property called transitivity. The precise statement and proof can be found in chapter 9 of M. Farber's book "Topology of closed one-forms". In what comes I am following Farber. That a closed one-form $\omega$ have non-degenerate zeros means that near each zero it can be written in the form $\omega = df$ with $f$ a Morse function. For such a one-form, the additional assumption of harmonicity means that the Morse index of a zero cannot be $0$ or $n$ (write $\omega = df$ near the zero; because $\omega$ is co-closed, $f$ is harmonic, so by the maximum principle cannot have a max or min at the zero). That $\omega$ be transitive means that for any point $p$ of $M$ which is not a zero of $\omega$ there is a smooth $\omega$-positive loop $\gamma: [0, 1] \to M$; that is, $\gamma(0) = p = \gamma(1)$, and $\omega(\dot{\gamma}(t)) > 0$ for $t \in [0, 1]$. Then Calabi's theorem states that a closed one-form with non-degenerate zeros is intrinsically harmonic if and only if it is transitive.
Near a non-degenerate index $0$ zero of a closed one-form the one-form can be written in the form $\delta_{ij}x^{i}dx^{j}$, for which it can be checked there are no positive loops beginning sufficiently near the origin.
(If one can handle $k$-forms then by Hodge duality one expects to be able to get somewhere with $(n-k)$-forms. The intrinsic harmonicity of $(n-1)$-forms was characterized in terms of transitivity in the thesis of Ko Honda, available on his web page). Evgeny Volkov (Volkov, Evgeny, Characterization of intrinsically harmonic forms, J. Topol. 1, No. 3, 643-650 (2008). ZBL1148.57036.)
weakens the non-degeneracy condition, replacing it with the condition that the closed one-form be locally intrinsically harmonic - that is, the restriction of the form to a suitable open neighborhood of its zero set is intrinsically harmonic.
As far as I know, for higher degree forms nothing much is known at all, though for some special cases, like $2$-forms on $4$-manifolds, something more has been said. One imagines that with further assumptions on the form, perhaps more can be said - for example a symplectic form is always intrinsically harmonic (use the metric determined by a compatible almost complex structure). On the other hand, Volkov's paper exhibits a closed $2$-form of rank $2$ on a $4$-manifold which is transitive but not intrinsically harmonic.
I am far from being an expert in this area, but I will try to present my understanding of this subject.
First of all, this is true that Hodge decomposition holds for smooth proper varieties over $\mathbf C$. However, the standard proof (that is explained below) uses the projective case as a black box.
Secondly, degeneration of Hodge-to-de Rham spectral sequence does imply that there is a filtration on $H^{n}(X,\mathbf C)$ s.t. the associated graded pieces are isomorphic to $H^{p,q}(X)$. In particular, it means that there is an abstract isomorphism of $\mathbf C$ vector spaces $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$. But I don't know any way to pick a canonical one just assuming degeneration of the spectral sequence.
Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the associated Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on the de Rham cohomology $H^n_{dR}(X)$ doesn't define a pure Hodge structure. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Hence the Hodge filtration doesn't define a pure Hodge structure on $H^1_{dR}(X)$ because of the failure of hodge symmetry. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first.
All of that being said, I will explain how to construct pure Hodge structures on cohomology of any proper smooth variety.
Let us start with $X$ being a smooth algebraic variety over $\mathbf C$. Then one can define a descending filtration on the algebraic de Rham complex $\Omega^{\bullet}_{X/\mathbf C}$ as $F^i\Omega^{\bullet}_{X/\mathbf C}=\Omega_{X/\mathbf C}^{\geq i}$. This induces a descending filtration $F^{\bullet}H^n_{dR}(X)=F^{\bullet}H^n(X,\mathbf C)$. The main result is that this construction defines a pure Hodge structure under the properness assumption on X.
Theorem 1: Let $X$ be a smooth and proper variety over $\mathbf C$, then a pair $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^{p}H^n(X,\mathbf C) \cap \overline{F^{n-p}H^n(X,\mathbf C)}\cong H^{q}(X,\Omega^p_{X/\mathbf C})$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$.
Remark 2: Although we have a canonical decomposition $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$, a priori there is no morphism $H^q(X,\Omega^p_{X/\mathbf C}) \to H^n(X,\mathbf C)$. I don't know any purely algebraic construction of these maps, which doesn't use the fact that Hodge filtration defines a pure Hodge structure as an input. I believe this is the crucial place where theory of harmonic forms comes into play. Once you have such morphisms, Theorem $1$ is basically equivalent to the degeneration of the Hodge-to-de Rham spectral sequence.
Remark 3: I think (I may be wrong) that there is no purely algebraic proof of Theorem $1$ as well. However, what I will show below is that if one takes Theorem 1 in the projective case as an input (this is the situation where $X^{an}$ is a Kahler manifold, so one can use theory of harmonic forms), then he/she can deduce the general form of Theorem $1$ in a purely algebraic way.
Before going to explain this proof, I need to mention a lemma which eventually be the key step to reduce a case of a proper variety to the case of a projective one.
Lemma 1: Let $f:X \to Y$ be a proper birational morphism between smooth connected varieties over a field $k$, then the induced morphism $H^p(f^*):H^p(Y,\Omega^q_{Y/k}) \to H^p(X,\Omega^q_{X/k})$ is injective for any $p$ and $q$.
Proof: Recall that $H^p(f^*)$ is induced by a morphism
$$
\Omega^q_{Y/k} \to \mathbf Rf_*f^*\Omega^q_{Y/k} \to \mathbf Rf_*\Omega^q_{X/k}.
$$
Compose it with the Trace map
$$
\operatorname{Tr_f}:\mathbf Rf_*\Omega^q_{X/k} \to \Omega^q_{Y/k}.
$$
(Here we use that $\operatorname{dim}X =\operatorname{dim}Y$ because in general the Trace map is map from $:\mathbf Rf_*\Omega^q_{X/k}$ to $\Omega^{q-d}_{Y/k}[-d]$ where $d=\operatorname{dim}X -\operatorname{dim}Y$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_f\circ f^*:\Omega^q_{Y/k} \to \Omega^q_{Y/k}$ is the identity morphism on an open dense subset $U\subset Y$ (because $f$ is birational and Y is normal!). Since $\Omega^q_{Y/k}$ is a locally free sheaf, we conclude that $Tr_f\circ f^*$ is also the identity morphism. Therefore, $H^p(Tr_f\circ f^*)=H^p(Tr_f)\circ H^p(f^*)$ is the identity morphism as well. So $H^p(f^*)$ is injective as desired.
Now let's come back to the proof of Theorem 1 in the proper case.
Step 1: Chow lemma and Hironaka's resolution of singularities allow us to find a smooth and projective variety $X'$ together with a map $f:X' \to X$ s.t. $f$ is birational and proper. Then there is a canonical morphisms of Hodge-to-de Rham spectral sequences for $X$ and $X'$. Namely, let
$$
E_1^{p,q}=H^q(X,\Omega^p_{X/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X).
$$
$$
\downarrow{f^*}
$$
$$
E_1'^{p,q}=H^q(X',\Omega^p_{X'/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X').
$$
Note that Lemma $1$ guarantees that $f^*$ is injective on the first page. Since we know that $E'$ degenerates on the first page ($d_1'^{p,q}=0$ for any $p,q$) we conclude that $d_1^{p,q}=0$ (because $f^*$ is compatible with differentials and injective). So $E_2^{p,q}=E_1^{p,q}$ and $E_2'^{p,q}=E_1'^{p,q}$, therefore $f^*$ is also injective on the second page. Then the same argument shows that $d_2^{p,q}=0$ for any $p,q$. Keep going to show that $E$ degenerates on the first page ($E_1^{p,q}=E_{\infty}^{p,q}$).
Step 2: Two consequences from the Step 1.
I) $F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)}=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^{\bullet}H^n_{dR}(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)} \subset F^pH^n_{dR}(X')\cap \overline{F^{n-p+1}H^n_{dR}(X')}=0.
$$
II) We have a canonical isomorphism $F^pH^n_{dR}(X)/F^{p+1}H^n_{dR}(X)\cong H^{q}(X,\Omega^p_{X/\mathbf C})$. This is a direct consequence of the degeneration of the Hodge-to-de Rham spectral sequence.
Step 3: The last thing we need to check is that $F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X)$.
Since $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}H^n_{dR}(X)}$ are disjoint inside $H^n_{dR}(X)$, it suffices to prove the equality
$$
\operatorname{dim}(F^pH^n_{dR}(X))+\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)})=\operatorname{dim}(H^n_{dR}(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
\operatorname{dim}(F^pH^n_{dR}(X))=\sum_{i\geq p}h^{i,n-i},
$$
$$\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)} = \sum_{i\geq n-p+1}h^{i,n-i},
$$
$$
\operatorname{dim}(H^n_{dR}(X))=\sum_{i\geq 0}h^{i,n-i} \text{, where } h^{p,q}= \operatorname{dim}H^q(X,\Omega^p_{X/\mathbf C}).
$$
Since we know that $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}}H^n_{dR}(X)$ are disjoint we conclude that
$$
\sum_{i\geq p}h^{i,n-i} +\sum_{i\geq n-p+1}h^{i,n-i} \leq \sum_{i\geq 0}h^{i,n-i}.
$$
Subtract $\sum_{i\geq n-p+1}h^{i,n-i}$ to obtain an inequality
$$
\sum_{i\geq p}h^{i,n-i} \leq \sum_{i\leq n-p}h^{i,n-i} (*)
$$
Our goal is to prove that this is actually an equality. Here comes the miracle! We use an additional symmetry between hodge numbers of interchange left and right hand sides of this inequality to actually show that it is an equality! Namely, we apply Serre duality to see that $h^{p,q}=h^{d-p,d-q}$ where $d=\operatorname{dim} X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
\sum_{i\geq p}h^{i,n-i}=\sum_{i\geq p}h^{d-i,d-n+i}=\sum_{j\leq d-p=(2d-n)-(d-n+p)} h^{j,(2d-n)-j},
$$
$$
\sum_{i\leq n-p}h^{i,n-i}=\sum_{i\leq n-p}h^{d-i,d-n+i}=\sum_{j\geq d-n+p}h^{j,2d-n-j}.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
\sum_{i\leq n-p}h^{i,n-i}=\sum_{j\geq d-n+p}h^{j,2d-n-j} \leq \sum_{j\geq d-n+p}h^{j,2d-n-j} =\sum_{i\leq n-p}h^{i,n-i}.
$$
Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$
F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X).
$$
In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.
Best Answer
There are no "rapid decaying harmonic 2-forms" in Minkowski space.
Consider the expression $$ 0 = \mathrm{d} \star \mathrm{d} F = \partial^i \partial_{[i}F_{jk]} = \frac13 (\Box F_{jk} + \partial_j \partial^i F_{ki} + \partial_k \partial^i F_{ij})$$ The second and third terms in the parentheses vanish by $\mathrm{d}\star F = 0$, so we see that over Minkowski space, Maxwell's equation implies that the components of the Faraday tensor must solve a linear wave equation.
(In the case you have source, you also see that the equations are equivalent to $\Box F_{jk} = J_{jk}$ where $J_{jk}$ is built out of the magnetic and electric currents in the appropriate way.)
Now, the linear decay estimates for the wave equation (that $|F| \sim \frac{1}{1+|t|}$) is sharp. In fact, by the finite speed of propagation + strong Huygens principle, if $F$ decays too rapidly in time, conservation of energy immediately implies that $F$ must vanish identically!
(If you don't want to go through the wave equation, you can also argue through the Maxwell equation by considering the energy integrals.)
Conservation of energy also implies that over Minkowski space $\mathbb{R}^{1,3}$ there cannot exist a non-vanishing "harmonic" form with finite space-time $L^2$ norm. So harmonic forms cannot exist, even in a class that is not necessary "rapidly" decaying.
Now, under the influence of external sources $j_e$ and $j_m$, you can pose the ansatz (by linearity of the equations) $F = G + H$, where $$ \begin{align} \mathrm{d} G & = j_e & \mathrm{d} H & = 0 \newline \mathrm{d} \star G & = 0 & \mathrm{d} \star H &= j_m \end{align} $$ Then the cohomology result you mentioned implies that $H = \mathrm{d}A$ and $\star G = \mathrm{d}C$ for $A,C$. The difference of the true solution with your ansatz is $\tilde{F} - F$ a harmonic two form, which can be completely absorbed into the electric potential $A$ if you want...
So the claimed decomposition is available whenever it is possible to solve the equations (possibly nonuniquely) for $G$ and $H$. For compatible source terms (you need $\mathrm{d} j_e =0 = \mathrm{d} j_m$ as a necessary condition) which are $L^1$ in space-time, the existence of a solution follows from the well-posedness of the corresponding Cauchy problem, for example.
Much of "Hodge theory" is false for "compact semi-Riemannian manifolds" in general. For example, there is no good connection between the cohomology and the number of harmonic forms. (Simplest example: on any compact Riemannian manifold the only harmonic functions are constants. But take, for example, the torus $\mathbb{T}^2 = \mathbb{S}^1\times\mathbb{S}^1$ with the Lorentzian metric $\mathrm{d}t^2 - \mathrm{d}x^2$. Then for any $k$ the function $\sin (kx) \sin(kt)$ is "harmonic", and they are linearly independent over $L^2$. )
Number 4 above also implies that the Hodge decomposition is not true for general compact semi-Riemannian manifolds. Consider the exact same $\mathbb{T}^2$ as above. Let $\alpha$ be the one form $\sin(t) \cos(x) \mathrm{d}t$. We have that $$ \mathrm{d}\alpha = \sin(t) \sin(x) \mathrm{d}t\wedge \mathrm{d}x \qquad \mathrm{d}\star \alpha = \cos(t) \cos(x) \mathrm{d}t \wedge \mathrm{d}x$$ To allow a Hodge decomposition requires that there exists $A$ and $C$, scalars, such that $$\mathrm{d}\star\mathrm{d}C = \mathrm{d}\alpha \implies \Box C = \pm \sin(t)\sin(x) $$ and $$ \mathrm{d}\star \mathrm{d}A = \mathrm{d}\star\alpha \implies \Box A = \pm \cos(t) \cos(x) $$ where $\Box$ is the wave operator $\partial_t^2 - \partial_x^2$.
Now taking Fourier transform (since we would desire $C$ and $A$ be in $L^2$ at least), we see that we have a problem. If $f\in L^2(\mathbb{T}^2)$ we have $$ \widehat{\Box f} = -(\tau^2 - \xi^2)\widehat{f} = 0 \qquad\text{ whenever } |\tau| = |\xi| $$ But the Fourier support of $\sin(t)\sin(x)$ and $\cos(t)\cos(x)$ are precisely when $\tau = \xi = 1$. So there in fact does not exist a Hodge-like decomposition for the $C^\infty$ form $\alpha$.
The above just goes to say that