I assume $u\in H^1(\Omega)$ for some nice domain $\Omega\subset\mathbb R^n$, and you wonder why $Lu\in H^{-1}(\Omega)$. (Correct me if I'm wrong.) For any $v\in H^1_0$, formal integration by parts gives $$ \langle Lu,v\rangle = \int_\Omega -a_{ij}D_juD_iv+b_iD_iub+cuv. $$ This integral makes perfect sense since $u,v,\nabla u,\nabla v\in L^2(\Omega)$. This is actually how one should define $L$ as an operator $L:H^1\to H^{-1}$, and it should not be hard show that it is continuous if the weights $a,b,c$ are good enough. The reason for taking $v\in H^1_0$ instead of $H^1$ is that it's nice to have a formulation without boundary terms that agrees with the classical one for smooth functions.

I assume $u\in H^1(\Omega)$ for some nice domain $\Omega\subset\mathbb R^n$, and you wonder why $Lu\in H^{-1}(\Omega)$. (Correct me if I'm wrong.) For any $v\in H^1_0$, formal integration by parts gives $$ \langle Lu,v\rangle = \int_\Omega -a_{ij}D_juD_iv+b_iD_iub+cuv. $$ This integral makes perfect sense since $u,v,\nabla u,\nabla v\in L^2(\Omega)$. This is actually how one should define $L$ as an operator $L:H^1\to H^{-1}$, and it should not be hard show that it is continuous if the weights $a,b,c$ are good enough. The reason for taking $v\in H^1_0$ instead of $H^1$ is that it's nice to have a formulation without boundary terms that agrees with the classical one for smooth functions.

Let me add explicitly that you can lose more derivatives than you have. If you have one derivative ($u\in H^1$), you lose two ($u\in H^{-1}$). You can define the operator $L$ between many spaces, not only $H^2\to H^0$. If you end up with $H^s$ with $s<0$, you just need to interpret the derivatives in a distributional sense.