This question is motivated by the test function spaces for distributions, but to be simple just let $K_i=[-i,i]\subseteq\mathbb R$, $i=1,2,\ldots$, and consider the inductive limit topology of the continuous functions $C_i$ on $\mathbb R$ with supports subsets of $K_i$, with the $\sup$ norm. Within the context of distribution theory, the usual definition for the topology of the continuous functions with compact support is the finest locally convex topology such that the inclusions are continuous, but here we take the straight inductive limit, dropping the locally convex condition.
For a monotone decreasing sequence $t$ of positive numbers $t_i$, define the set $U_t$ to be the continuous functions of compact support with supremum on $K_i$ smaller than $t_i$. It is easily seen that the sets $U_t$ are convex. Claim: the collection of $U_t$ is an neighborhood base at $0$ in the inductive limit topology. $U_t\cap C_i$ is the ball of functions in $C_i$ with supremum smaller than $t_i$, and so $U_t$ is open in the inductive limit topology. Suppose $U\ni 0$ is open. Then $U\cap C_i$ contains a ball about $0$, say of radius $t_i$. Inductively, one can choose $t_i$ decreasing. Let $f\in U_t$ have compact support $K$, and choose $i$ so large that $K\subseteq[-i,i]$. Then $f$ has supremum smaller than $t_i$ and support within $K_i$, so $f\in U$, because that ball is contained in $U$. Thus the straight inductive limit topology is already locally convex in this case.
The topologies used for distributional test function spaces take steps (convex hulls) to manually ensure the relevant inductive limit topologies are locally convex. They must have to do that. But the above seems to indicate that those spaces are locally convex apriori. No need for the convex hulls. What's wrong here?
Added in edit: The above argument also shows that the inductive limit of normed spaces is locally convex. But the spaces $D_{K_i}$ are not normed. The analogous statement for that case has for a monotone sequence $t=\{t_i\}$ and an increasing sequence $n=\{n_i\}$, the sets $U_{t,n}$, with derivatives up to an including $n_i$ bounded in absolute value by $t_i$ on $K_i$, are a base for the inductive limit topology. These sorts of sets are inherently nonconvex, because a convex sum of two functions may increase the support of either, and so, given membership in $U$, such a convex sum may increase the control on higher derivatives around $0$, but simultaneously may not comply and decrease such derivatives accordingly.
For example, take a function $f$ with support in $[1/4,3/4]$ with maximum $1$ but high first derivative. This can be in $U_{t,n}$ with $t_1=1$ and $n_1=0$. Add to this a function $g$ with support in $[5/4,7/4]$ which has value and first derivative less than $1/2$ in absolute value. This can be in $U_{t,n}$ with $t_2=1/2$ and $n_2=1$. But $(f+g)/2$ may not be in such a $U_{t,n}$ because the support of that is not within $[-1,1]$ and support exceeding $[-1,1]$ has first derivative smaller than $1/2$, which need not be the case for $(f+g)/2$.
The answer to the question posed is: nothing is wrong, but the argument extended does not lead to a convex local base at $0$ for the inductive limit topology on the space of distributional test functions.
Such explicit local bases seem helpful for understanding. For example, even in the continuous case in the question originally posed, one can easily form a net on the sequences $t$, by translating a bump function $f_0$ with support in, say, $[1/4,3/4]$ by $i$ and multiplying by $t_i$. That is, $f_t(x)=t_i f_0(x-i)$ if $x\in K_i$. Such a net does not have support in any $K_i$ but it does converge to the zero function.
The argument in https://mathoverflow.net/a/333618/59219 seems to show that if $U$ is a convex open set in the inductive limit topology and $U_{t,n}\subseteq U$ then the sequence $n$ is eventually constant. This would seem to imply that the $U_{t,n}$ with $n$ a constant sequence, is a local base at $0$ for the test function topology, which would in my opinion clarify that topology, by providing an easy to understand local base at $0$.