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 $0$.
Your proof is correct. However, it is easier to argue that $\{ x \in X : f(x) g(x) \neq 0 \} \subset \{ x \in X : f(x) \neq 0 \}$ which implies $\text{supp} (f \cdot g) \subset \text{supp} (f)$: Hence $\text{supp} (f \cdot g)$ is compact because closed subsets of compact sets are compact.
Therefore $K_c(X)$ is a $K$-algebra for any topological field $K$.
Best Answer
A space $X$ is $\sigma$-compact if it can be written as the countable union of compact subsets. Therefore one can find a sequence of compact subsets $K_n$ of $X$ with $K_n \subset K_{n+1}$ and $\bigcup K_n =X$. However, we do not necessarily find a sequence of compact subsets $K_n$ of $X$ with $K_n \subset \text{int}(K_{n+1})$ and $\bigcup K_n =X$. Call such a sequence strongly ascending. Clearly, the existence of a strongly ascending sequence of compact subsets implies $\sigma$-compactness.
Note that there is a subtle point concerning the notion of compactness. Some authors understand this to include Hausdorff, other authors do not. In my answer I shall not assume it.
Let us now define a space $X$ to be locally compact if each $x \in X$ has an open neighborhood whose closure is compact and to be strongly locally compact if for each $x \in X$ and each open neighborhood $U$ of $x$ there exists an open neighborhood $V$ of $x$ such that $\overline V$ is compact and $\overline V \subset U$. In my opinion the latter is the better concept, but there is no real standard. Anyway, if $X$ is Hausdorff, then both concepts agree. An example where they differ is the Alexandroff compactification $\alpha(\mathbb Q)$ of the rationals $\mathbb Q$. This space is compact, hence locally compact, but not strongly locally compact. See my answer to An example of a compact topological space which is not the continuous image of a compact Hausdorff space?.
Let us prove
Lemma. $X$ admits a strongly ascending sequence of compact subsets if and only if $X$ is $\sigma$-compact and locally compact.
Proof. (1) Let $X$ admit a strongly ascending sequence of compact subsets $K_n$. Then $X = \bigcup \text{int}K_{n+1}$, hence each $x \in X$ is contained in some $\text{int}K_{n+1}$ so that $K_{n+1}$ is a compact neigborborhood of $x$. Therefore $X$ is locally compact.
(2) Let $X$ is $\sigma$-compact and locally compact. Write $X= \bigcup C_n$ with $C_n$ compact. By induction we construct a sequence of compact $K_n \subset X$ such that $$(*) \quad K_i \cup C_i \subset \text{int}(K_{i+1})$$ for all $i$. Then $(K_n)$ is strongly ascending.
For $n=1$ let $K_1 = C_1$. Assume we have found $K_1,\ldots, K_n$ such that $(*)$ is satisfied for all $i < n$. We now construct $K_{n+1}$. By local compactness each $x \in K_n \cup C_n$ has an open neighborhood $U(x)$ such that $D(x) =\overline{U(x)}$ is compact. Since $K_n \cup C_n$ is compact, there are finitely many $x_i \in K_n \cup C_n$ such that $K_n \cup C_n \subset V = \bigcup_{i=1}^r U(x_i)$. Let $K_{n+1} = \bigcup_{i=1}^r D(x_i)$. This is a compact set such that $K_n \cup C_n \subset V \subset K_{n+1}$. But since $V$ is open, we get $K_n \cup C_n \subset \text{int}(K_{n+1})$.
For your purposes you need a strongly ascending sequence of compact subsets. To apply Urysohn's Lemma, you must assume that the $K_n$ are closed and normal. This is automatically satisfied if $X$ is Hausdorff. If it is not, then you get fairly technical assumptions concerning a very special strongly ascending sequence.