I will try to start from the notion of support of a function and obtain the definitions above in a natural way.
If $f : \mathbb{R}^n \to R$ then its support is defined as $S = \overline{\{x \in \mathbb{R}^n : f(x) \neq 0\}} $ For the purpose of discussion it's easier to talk about $S^c$ instead of $S$, namely $S^c$ is the largest open set where $f = 0$.
So far, so good, but distributions are not functions, so it doesn't make sense to say that the value of a distribution at a point is $0, -1, \pi$, etc. However, distributions are linear functionals, so it's not unreasonable to define that a distribution $T$ is zero on an open set $\omega$ if it "doesn't do anything there". In other words, for an arbitrary $\phi$ smooth, compactly supported in $\omega$ then $\langle T, \phi \rangle = 0$. Thus, we have arrived at the definition of open annihilation set that you mentioned.
Now, to define the support of $T$ we take the complement of the largest open set where $T$ vanishes: just like in the case for support of a function $f$: look at the disussion about $S$ and $S^c$ above.
I hope this helps.
Note: it's worth checking that if $T$ is induced by a (locally) integrable function $f$ in the standard way, then the support of $T$ will be the support of $f$, in other words the two definitions are consistent.
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
It is the (vector) of distributions defined via $$\nabla p : D(\Omega)^n \to \mathbb{R}, \quad \varphi \mapsto -\sum_{i=1}^np(\partial_{x_i} \varphi_i).$$
If $p$ is actually a funcion, the last sum is just $$-\int_\Omega p(x) \, \operatorname{div}(\varphi)(x) \, \mathrm{d}x,$$ and if $p$ is (weakly) differentiable, you have $$-\int_\Omega p(x) \, \operatorname{div}(\varphi)(x) \, \mathrm{d}x = \int_\Omega \nabla p(x) \cdot \varphi(x) \, \mathrm{d}x.$$