On the one hand, Eskin's book on PDEs tells me that I should be content to think of this topology as one "described" (not fully, and it's not even clear it's a topology) by the convergence of sequences. That is that $f_n\rightarrow f$ iff $f_n$ have all their supports contained in one big compact $K$ and on $K$ we have sup norm convergence of $f_n$ and all its derivatives to the corresponding derivatives of $f$.
If I upgrade this definition to a convergence of nets in the same way, then at least it could be something that would fully describe a topology, not that all demands of how nets converge can be fulfilled by some topology. Could you please tell me why this set of requirements of convergence of nets is consistent with not only some topology, but the one resulting from the inductive limit construction?
To remind you, that construction proceeds very technically, and then the characterizing properties are defined. We have that all the inclusions from $C^\infty_K$ (Supports are a subset of $K$) to $C^\infty_c(\mathbb{R}^d)$ are continuous, and that any linear map from $C^\infty_c(\mathbb{R}^d)$ to some locally convex topological vector space is continuous just as soon as its compositions with all the inclusion maps are continuous.
The construction for the locally convex inductive limit topology on $C^\infty_c$ is not technical at all: It is the locally convex tology generated by all semi-norms $p$ all whose restrictions to $C^\infty_K$ are continuous.
However, the statement you claim is not true: There are many nets which converge to $0$ without having support in a fixed compact set. This is most easily seen in an abstract setting where $X_n \subseteq X_{n+1}$ is any properly increasing sequence of locally convex spaces: Consider the $0$-neighborhood filter $\mathcal U$ of the inductive limit $X_\infty$ directed by $U \le V$ if $V\subseteq U$ (sic). Then take $I=\mathcal U \times \mathbb N$ (directed in the obvious way) as an index set and for $i=(U,n) \in I$ choose $x_i \in U \setminus X_n$ (this is possible since $U$ is absorbing in $X_\infty$ which is not contained in $X_n$). Then $(x_i)_{i\in I}$ is a net which converges to $0$ without being contained in any $X_n$.
EDIT.
That locally convex topologies can be described either by systems of absolutely convex absorbing sets ($0$-neighborhoods) or, equivalently, by systems of semi-norms is not very difficult: Each semi-norm is essentially described by its unit ball and vice versa, each absorbing absolutely convex set gives a semi-norm by its Minkowski functional.
That convergent sequences of the inductive limit are contained and convergent in some step (this property is called sequentially retractive) needs stronger assumptions, e.g., that each $X_n$ is a closed topological subspace of some Hausdorff locally convex space $Y$ (which is the case for $\mathscr D$ since $C^\infty_K$ are closed in $C^\infty$). The proof is by contradiction: Consider a sequence $x_n$ in $X_\infty$ which converges to $0$ and satisfies $x_n\notin X_n$ for all $n$. Since $X_n$ is closed in $Y$ you find absolutely convex $0$-neighborhoods $U_n$ in $Y$ such that $x_n\notin X_n+U_n$. Then you check that $U=\bigcap_n (X_n+U_n)$ is a $0$-neighborhood in $X_\infty$ (because $U\cap X_n$ is a $0$-neighborhood in $X_n$ for each $n$). As $x_n\to 0$ you get $x_n \in U$ for all but finitely many $n$ which yields the contradiction $x_n\in U\subseteq X_n+U_n$. We have shown that every null sequence in $X_\infty$ is contained in some step $X_n$. Since the sequence converges to $0$ in $Y$ it then converges also in $X_n$.
Although this is not very difficult, such arguments are (in my opinion, unfortunately) missing in almost all textbooks on functional analysis.