Intuitive explanation for $P(A \cap B) - P(A)P(B) = P(\bar{A} \cap \bar{B}) - P(\bar{A}) P(\bar{B})$ For two events $A$, $B$ and a probability measure $P$, one has
\begin{equation*}
  P(A \cap B) - P(A)P(B) = P(\bar{A} \cap \bar{B}) - P(\bar{A}) P(\bar{B}),
\end{equation*}
where the bars denote complements. The probability measure is not special as the identity remains true for measurable sets of an arbitrary measure (including infinite ones).
The proof by direct calculation is of course easy and elementary, what I'm looking for is an intuitive, hopefully somewhat visual argument, in particular clearly describing the inherent symmetry.
 A: A small caveat on notation
I'll be using $S^c$ to denote the complement of an event $S$, instead of $\bar{S}$. The bar notation will be used in one of the sections later on.
A convenient rewrite
One way of thinking about this inequality that I like would be to write it as $$
P(A \cap B) - P(A)P(B) = E[1_{A}1_B] - E[1_A]E[1_B]
$$
where $$
1_X(\omega) = \begin{cases}
1 & \omega \in X\\
0 & \omega \notin X
\end{cases}
$$
is the indicator function of an event $X$ which is a subset of a sample space $\Omega$. The equality given follows from $1_S1_{S'} = 1_{S \cap S'}$ and $E[1_D] = P(D)$ for any events $S,S',D$.
Recognize this as the covariance $$
\mbox{Cov}(1_A,1_B) = E[1_A1_B] - E[1_A]E[1_B] = E[(1_A - P(A))(1_B-P(B))]
$$
We see that we are basically comparing the covariances of $1_A$ and $1_B$. In the other, we are comparing the covariances of $1_{A^c}$ and $1_{B^c}$.

Probabilistic interpretation of covariance
Now, what are some interpretations of covariance? From a probabilistic interpretation, we have the identity $$
1_A - P(A) = P(A^c) - 1_{A^c} = -(1_{A^c} - P(A^c))
$$
Now, $1_A - P(A)$ is basically $1_A - E[1_A]$ , the "fluctuation from the mean" of the event $1_A$. On the other hand, $1_{A^c} - P(A^c)$ is the "fluctuation from the mean" of the event $1_{A^c}$.
The identity above states that the fluctuation from $1_A$ from its mean is "opposite" to the fluctuation of $1_{A^c}$ from its mean, for any event $A$. That is, when $1_A$ is "above" its mean, $1_{A^c}$ is "below" its mean, and this is an exact identity as functions on $\Omega$. The same will be true for the event $B$ as well.
Now, covariance is a measure of how random variables "jointly" vary. Roughly, the covariance of $A$ and $B$ is high if , whenever $A$ tends to be above its mean, then $B$ also tends to be above its mean (and vice-versa) as random variables. Similarly, the covariance is negative if , whenever $A$ tends to be above its mean, $B$ tends to be below its mean (and vice-versa). The covariance is zero if the the fluctuations between the two random variables are uncorrelated (but not independent), and its magnitude alone will dictate the amount $A$ and $B$ are correlated.
Once you use the identity, you recognize immediately that $$
\mbox{Cov}(1_A,1_B)=E[(1_A - P(A))(1_B-P(B))] = E[\{-(1_{A^c} - P(A^c)\}\{-(1_{B^c} - P(B^c)\}] \\= E[(1_{A^c} - P(A^c))(1_{B^c} - P(B^c))] = \mbox{Cov}(1_{A^c},1_{B^c})
$$
which then leads to the result. As a one-liner, the best way of putting this is : the correlation between the fluctuations of $1_A$ and $1_B$ around their mean, is exactly equal to the correlation between the fluctuations of $1_{A^c}$ and $1_{B^c}$ around their mean.
What you've actually seen is often called a "sample element" argument. This is an argument that is often sought after in probability theory, for explaining probabilistic equivalences, and basically explains the equality by saying "two functions are equal, therefore their expectations are equal".
That is, the amount that $1_A$ varies from its mean, is exactly the same magnitude, but opposite in sign to that of $1_{A^c}$. The same applies to $1_{B^c}$. Therefore, multiplying these fluctuations, the product of the fluctuations is actually the same for each $\omega \in \Omega$, therefore the product of the fluctuations are equal as random variables, and therefore have the same expectation.

Geometric Interpretation of Covariance
The geometric interpretation of covariance is as a (semi) inner product. Allow me some time to explain this.
Let $\Omega$ be a finite sample space, $\Omega = \{x_1,x_2,\ldots,x_n\}$.  Let $X$ be a real valued random variable on $\Omega$, and look at the centered random variable $\bar{X} = X - E[X]$( a centered random variable has mean zero). How can we form a "vector" here? The answer is to literally look at the point $(X(x_1),X(x_2),\ldots,X(x_n)) \in \mathbb R^n$. This is the exact description of $X$ : looking at a point in $\mathbb R^n$, you instantly know the function it corresponds to, and vice-versa.
Now, let's take $1_A$ and $1_{A^c}$. What can we say about the vectors corresponding to the random variables $\overline{1_A}$ and $\overline{1_{A^c}}$? Since we have $$
1_A - P(A) = P(A^c) - 1_{A^c} = -(1_{A^c} - P(A^c)) \implies \overline{1_A} = - \overline{1_{A^c}}
$$
The vectors corresponding to $1_A$ and $1_{A^c}$ are of the same magnitude and point in exactly opposite directions! The same would be true for $1_B$ and $1_{B^c}$.
Now, watch what happens when we interpret centered random variables as points.

Given centered random variables $\overline{X},\overline{Y}$, the quantity $\mbox{Cov}(X,Y)$ is actually equal to the dot product of the vectors corresponding to $X$ and $Y$!

To give a rough idea of this, because $E[X],E[Y] = 0$, the dot product gives an idea of how much the vectors corresponding to $X$ and $Y$ "point and increase in similar directions", and the covariance does the same job as well.
Now, imagine you have two vectors $\vec{u},\vec{v}\in \mathbb R^m$. It is fairly obvious that
$$
\vec{u} \cdot \vec{v} = (-\vec{u}) \cdot (- \vec{v})
$$
This is because the angle between $-\vec{u}$ and $-\vec{v}$ is the same as that between $\vec{u}$ and $\vec{v}$, simply by inspection (or , more formally, by using the linear transformations that flips the direction of every axis to travel between the two inner products, and noting that such transformations are "orthogonal" i.e. preserve angles between vectors).
Once we translate the inner product equality above to the probability setting using $\vec{u} = \bar{1_A}$ and $\vec{v} = \overline{1_B}$, we see that we get back the covariance equality and have another interpretation of the identity.
