I can't comment yet so let me post as answer. Forget about the coefficients and just take $L=-\Delta$, the Dirichlet Laplacian. Then you are asking why $-\Delta u \in H^{-1}(\Omega)$ when $u \in H^1_0(\Omega)$. It is because the weak Laplacian is defined $$\langle -\Delta u, v \rangle_{H^{-1}, H^1_0} := \int_\Omega \nabla u \nabla v.$$ It is easy to verify that this operator $-\Delta u$ is a bounded linear functional belonging to the dual space of $H^1_0$.

So it seems incorrect in your OP to say " I can't figure out this simple fact." because it is a definition.

I can't comment yet so let me post as answer. Forget about the coefficients and just take $L=-\Delta$, the Dirichlet Laplacian. Then you are asking why $-\Delta u \in H^{-1}(\Omega)$ when $u \in H^1_0(\Omega)$. It is because the weak Laplacian is defined $$\langle -\Delta u, v \rangle_{H^{-1}, H^1_0} := \int_\Omega \nabla u \nabla v.$$ It is easy to verify that this operator $-\Delta u$ is a bounded linear functional belonging to the dual space of $H^1_0$.

So it seems incorrect in your OP to say " I can't figure out this simple fact." because it is a definition.

When you say that $-\Delta u =f$ holds as an equality in $H^{-1}$, it means exactly that $$\int_\Omega \nabla u \nabla v = \langle f, v \rangle_{H^{-1}, H^1_0}$$ holds for all $v \in H^1_0$. When you say that $-\Delta u =f$ holds as equality in $L^2$, you are saying that in fact $-\Delta u$ is in the subspace $L^2 \subset H^{-1}$, and therefore in addition to the above, $-\Delta u(x) = f(x)$ holds pointwise a.e. $x \in \Omega$.

I think this probably is not the meaning you intended (you probably meant the weak formulation holds but the right hand side becomes $\int_\Omega fv$ since $f \in L^2$) but you should be aware of it.