First question: is $E$ equipped with this norm a Banach space?
No. In fact, its completion is the space of continuous functions $\mathbb{R}^n\to\mathbb{R}^{m}$ that vanish at $\infty$ (that is, $f(x)\to 0$ as $\|x\|\to\infty$). Given any continuous function $f$ which vanishes at $\infty$, let $f_k$ be obtained by "cutting off" $f$ to be supported on a ball of radius $k$ around the origin (so, say, $f_k$ agrees with $f$ on a ball of radius $k-1$, and interpolates radially to $0$ between the sphere of radius $k-1$ and the sphere of radius $k$). Then $f_k\in E$ for each $k$ and $f_k$ converges uniformly to $f$, and it follows easily that $(f_k)$ is Cauchy in $E$ but has no limit in $E$ unless $f$ has compact support.
Second question: does this notion of convergence somehow generates a topology on $E$? Is it metrizable?
There is no metrizable topology which induces this notion of convergence. To prove this, for each $R$, choose a sequence $(\varphi_k^R)_k$ such that each $\varphi_k^R$ has a ball of radius $R$ as its support but $\varphi_k^R\to 0$ uniformly. Then each of these sequences $(\varphi_k^R)$ converges to $0$ in Maggi's sense. If this notion of convergence came from a metric, we could choose for each $R$ a $k_R$ such that $\varphi_{k_R}^R$ is within $1/R$ of $0$ with respect to the metric, and then $\varphi_{k_R}^R$ would converge to $0$ as $R\to\infty$. But this is impossible since convergent sequences are required to have some uniform compact support.
However, this convergence does come from a natural topology. For each compact $K\subset\mathbb{R}^n$, let $E_K$ be the subspace of $E$ consisting of functions with support contained in $K$. Say that a set $U\subseteq E$ is a basic neighborhood of $0$ if $U$ is balanced and convex, and $U\cap E_K$ is an open neighborhood of $0$ in $E_K$ with respect to the sup norm on $E_K$ for each $K$. Finally, say that a set $U\subseteq E$ is open if it is a union of translates of basic neighborhoods of $0$.
It can be shown that this topology makes $E$ as the colimit of the spaces $E_K$ with their sup norms in the category of locally convex topological vector spaces. That is, it is the finest topology on $E$ which makes the inclusion maps $E_K\to E$ all continuous and makes $E$ a locally convex topological vector space. It is then clear that any sequence which converges in Maggi's sense converges in this topology, since such a sequence will converge in $E_K$ with respect to the sup norm.
Proving the converse (every sequence convergent in this topology converges in Maggi's sense) is a little messier, but here's the idea. Suppose $(f_k)$ is a sequence of functions in $E$ such that the supports of the $f_k$ are not contained in any fixed compact set. Then, passing to a subsequence of $(f_k)$ if necessary, we can find a sequence $(x_k)$ going to $\infty$ in $\mathbb{R}^n$ such that $f_k(x_k)\neq 0$ for each $k$. Now let $U$ be the set of $f\in E$ such that $|f(x_k)|<|f_k(x_k)|$ for each $k$. Then $U$ is a basic neighborhood of $0$ in $E$ (this uses the fact that any $K$ contains only finitely many of the $x_k$). Since $f_k\not\in U$ for all $k$, this means $(f_k)$ cannot converge to $0$.
$\phi_{-Df}(g)=\int_\mathbb{R}-Dfgdx=\int_\mathbb{R}D(fg)dx-\int_\mathbb{R}Dfgdx=$
$=\int_\mathbb{R}fD(g)dx=\phi_f(Dg)$
While if $D’\phi_f=0$ then
$\phi_{-Df}(g)=-\int_\mathbb{R}D(f)gdx=0$
for each $g$
In general when you have a $C^\infty_c(\mathbb{R})$ function $h$ such that
$\int_\mathbb{R}hgdx=0$ for each $g$ then
$h=0$
In fact you can choose $g=h$ to have
$\int_\mathbb{R}h^2dx=0$ that implies $h=0$
So in your case $D(f)=0$ that means $f=c$
You can observe that $c=0$ because $f$ is continuos with compact support.
Best Answer
By dominated converge the space of compactly supported $L_\Delta^2(\mathbb{R})$ functions are dense in $L_\Delta^2(\mathbb{R})$ (you can approximate $f$ by the sequence $f_n= f 1_{[-n,n]}$). Using that $$\Vert g \Vert_{L^2([-n,n])} \leq \Vert g\Vert_{L_\Delta^2([-n,n])} \leq \sqrt{n^2+1} \Vert g\Vert_{L^2([-n,n])}$$ and the fact that $C_c(\mathbb{R})$ is dense in $L^2(\mathbb{R})$, we conclude that $C_c(\mathbb{R})$ is dense in $L_\Delta^2(\mathbb{R})$.