How is $P(A^c \cap B^c)$ the same as $1-P(A \cup B)$?

Your first diagram is incorrect. You seem to have drawn $A^c\cup B^c$ (as some other people have mentioned).

I would recommend drawing $A^c$ independently first, which consists of the rest of the universe and $B-A$. Then draw $B^c$ in a different color, noting again that you have the rest of the universe and $A-B$. The intersection of $A-B$ and $B-A$ is neither $A$ nor $B$, so you will end up without $A$ or $B$ in your final intersection. However, the rest of the universe is in both $A^c$ and $B^c$, therefore so is its intersection. Thus, you get that $P(A^c\cap B^c)$ is just the universe without $A$ or $B$, which is equivalent to $1-P(A\cup B)$.

According to the DeMorgan's laws, one has \begin{align*} \textbf{P}(\Omega) & = \textbf{P}((A\cup B)\cup(A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}((A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}(A^{c}\cap B^{c}) = 1 \end{align*} from whence the result follows immediately, since $X\cup X^{c} = \Omega$ and $X\cap X^{c} = \varnothing$ for every event $X\subseteq\Omega$.