Why Gateaux derivative is a distribution? Thanks to Jan Bohr answer and comment I edited this question.
Let $E$  be a vector bundle , $E^*$ the dual bundle and $D$ a density bundle. Denote by $\Gamma(E)$ the space of section of the bundle $E$.
By definition the a distribution $\omega$ in  a vector bundle $E$ is $\omega \in\Gamma(E^*\otimes D)^\prime$ where  $\Gamma(E^*\otimes D)^\prime$ is the topological dual of $\Gamma(E^*\otimes D)$.
Now in this paper Algebraic Structure of Classical Field Theory: Kinematics and Linearized Dynamics for Real Scalar Fields  by  Romeo Brunetti, Klaus Fredenhagen, Pedro Lauridsen Ribeiro they define at page 14 the Gateaux derivative as $$F^K[\phi](\phi_1...,\phi_k)=\left<F^K[\phi],\phi_1\otimes...\otimes\phi_k\right>=\frac{d^k}{d\lambda_1...d\lambda_k }\left(\phi + \lambda_1\phi_1+...\lambda_k\phi_k\right)$$
where $<>$ denotes dual pairing and $\phi,\phi_1...,\phi_k \in\Gamma(E)  $
Now they say that $F^K[\phi]$ is a distribuition . But  $F^K[\phi]$ is not in  $\Gamma(E^*\otimes D)^\prime$ how can it be a distribution?
Thanks to Jan Bohr answer and comment I add this part .
In this paper Properties of field functionals and characterization of local functionals by Christian Brouder, Nguyen Viet Dang, Camille Laurent-Gengoux, Kasia Rejzner at page 10
they define $F^K[\phi]$ as a d distributional section on $M^k$ for a general vector bundle.
Maybe they are using $<>$ to identify $E$ with $E^*$ but I am not sure
 A: Consider the following example: $M$ is a closed manifold, $\mu$ is a volume form and $F\colon C^\infty(M)\rightarrow \mathbb C$ is defined by $F(\varphi)=\int_M\varphi^2\mu$. Then $F(\varphi+\lambda\varphi_1)=F(\varphi)+2\lambda\int_M(\varphi\varphi_1)\mu+\lambda^2F(\varphi_1)$. Taking the $\lambda$-derivative at $0$, you obtain $$
dF_\varphi(\varphi_1)
=\int_M\varphi_1 (2\varphi\mu).
$$
For fixed $\varphi$ you can interpret this as the linear functional on $C^\infty(M)$ that takes some $\varphi_1$ and sends it to the integral agains the form $2\varphi\mu$. If we define  $\mathcal D'(M)$ as the continuous dual of $C^\infty(M)$, we have all right to call $dF_\varphi$ a distribution. As you mentioned, sometimes $\mathcal D'(M)$ is defined as the dual of $\Gamma(M,D)$, which is convenient if one does not want to fix a particular density. However, once a volume density is chosen (e.g. $\vert \mu \vert$), there is an isomorphism $C^\infty(M)\rightarrow \Gamma(M,D)$ given by $\psi\mapsto \psi \vert \mu \vert$. Hence also the dual spaces are isomorphic and you can consider $dF_\varphi$ as a distribution also in the other sense.
The authors write $dF_\varphi(\varphi_1)$ as $F^{(1)}[\varphi](\varphi_1)$ and then also consider higher order derivatives $F^{(k)}[\varphi](\varphi_1,\dots,\varphi_k)$. Again for fixed $\varphi$ you can view this as a linear functional on the algebraic tensor product $C^\infty(M)\otimes \dots \otimes C^\infty(M)$ ($k$-times). After choosing the correct topological tensor product, this should extend to a continuous linear functional on $C^\infty(M\times\dots\times M)$ and hence a distribution on $M^k$.
