Finally, I found the answers to my two questions, and they are both positive as I conjectured.
Take a sequence of compact sets $(K_m)_{m=0}^{\infty}$ in $\Omega$, each one with nonempty interior, such that:
(i) $K_m$ is contained in the interior of $K_{m+1}$ for each $m=0,1,\dots$;
(ii) $\cup_{m=0}^{\infty} K_m = \Omega$.
Let $(x_m)_{m=0}^{\infty}$ be a sequence in $\Omega$ such that $x_m$ lies in the interior of $K_m$ and $x_m \notin K_{m-1}$ (with $K_{-1}=\emptyset$). Define the set
\begin{equation}
V = \{ f \in \mathcal{D}(\Omega) : \left| f(x_{|\alpha|}) D^{\alpha} f(x_0) \right| < 1, | \alpha |=0,1,2, \dots \}.
\end{equation}
Let $K \subseteq \Omega$ be a compact set. Since only finitely many of the $x_m$'s belong to $K$, it is immediate to see that $V \cap \mathcal{D}_K \in \tau_K$. Assume that $V$ contains some $\tau$-open set containing 0. Then, since $\mathcal{D}(\Omega)$ with the topology $\tau$ is a locally convex topological vector space, there would exist a convex balanced set $W \subseteq V$ such that $W \in \tau$. So $W \cap \mathcal{D}_K \in \tau_K$ for each compact $K \subseteq \Omega$.
Then for each m, there exists a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set
\begin{equation}
U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \}
\end{equation}
is contained in $W \cap \mathcal{D}_{K_m}$. Let $m=N(0)+1$. Then the interior of $K_m$ contains the point $x_{N(0)+1}$, so that there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $cf+(1-c)g$ does not belong to $V$. So $W$ is not convex, against the hypothesis. This shows that (Q2), and so (Q1), has a positive answer.
NOTE (1). Actually, this example also shows that $\mathcal{D}(\Omega)$ with the topology $\tau'$ is not even a topological vector space. Indeed, if it were, then we should be able to find $S \in \tau'$ such that $S + S \subseteq V$. Again, we could find then for each $m$ a positive integer $P(m)$ and $\delta(m) > 0$ such that the set
\begin{equation}
T_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{P(m)} < \delta(m) \right \}
\end{equation}
is contained in $S \cap \mathcal{D}_{K_m}$. Choose $m=P(0)+1$, so that the interior of $K_m$ contains $x_{P(0)+1}$. Then there exists $g \in T_m$ such that $|g(x_{P(0)+1})| > 0$. As before, note for any $M > 0$, we can find $f \in T_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = P(0)+1$. This in turn implies that there exists $f \in T_0$ such that $f+g \notin V$.
QED
NOTE (2). We can prove in the same way as before that for every $f \in V$, there is no $U \in \tau$ such that $f \in U$ and $U \subseteq V$. Assume there exists. Then, being $\mathcal{D}(\Omega)$ with the topology $\tau$ a locally convex space, we can find a convex balanced set $W \in \tau$ such that $f + W \subseteq U$. Again, we could find then for each $m$ a positive integer $N(m)$ and $\epsilon(m) > 0$ such that the set
\begin{equation}
U_m = \left \{ f \in \mathcal{D}_{K_m}: \left| \left|f \right| \right|_{N(m)} < \epsilon(m) \right \}
\end{equation}
is contained in $W \cap \mathcal{D}_{K_m}$. Choose then $m=N(0)+1$, so that the interior of $K_m$ contains the point $x_{N(0)+1}$. Then there exists $g \in U_m$ such that $|g(x_{N(0)+1})| > 0$ and $|g(x_{N(0)+1})| < | \varphi(x_{N(0)+1})|$ if $| \varphi(x_{N(0)+1})| > 0$. Now note that for any $M > 0$, we can find $f \in U_0$ such that $|D^{\alpha}f(x_0)| > M$ for some multi-index $\alpha$ such that $|\alpha| = N(0)+1$. This in turn implies that for any $c \in (0,1)$, we can find $f \in U_0$ such that $\varphi + cf+(1-c)g$ does not belong to $V$, which gives a contradiction, since $cf+(1-c)g \in W$.
Best Answer
OK, I think I got a counterexample now, i.e. the final topology w.r.t. to the inclusion maps is not linear. I only worked it through on ${\bf R}$ though.
The only assumption that I need is that for given $\epsilon>0$ and $k,n\in{\bf N}$ we find some $f\in C^\infty([-1,1])$ such that $\|f^{(j)}\|_\infty<\epsilon$ for $j<k$ and $f^{(k)}(0)>n$, which I think is easily constructable.
Now the set $$A:=\{f\in C^\infty_K:\forall j>0.f(j)<1/f^{(j)}(0)\}$$ (where we set $1/0=\infty$) is open in the final topology, i.e. $A\cap C^\infty([-i,i])$ is open for each $i$. Yet, if the final topology were linear, we would find an neighbourhood $B$ of $0$ such that $B+B\subseteq A$. But any such $B$ contains a basic open neighbourhood from $C^\infty([-i,i])$ of the form $$U_i=\{f\in C^\infty([-i,i]):\forall j<k(i).\|f^{(j)}\|_\infty<\epsilon(i)\}$$ for each $i$. Now we can construct $f\in U_1$ such that $f^{(k(i))}(0)$ is large enough so $f+U_{k(i)+1}\subseteq B+B$ is no longer a subset of $A$.
So the majority of books and scripts I read (e.g. Rudin) got it right, when they defined the topology on the space of test functions as the limit topology in the category of locally convex spaces, i.e. the finest locally convex topology which makes the inclusion maps continuous.