Sorry but the analysis of what commutators mean (in the given link) although very good, does not provide intuition and does not generalise to anti-commutators.
Commutators used for Bose particles make the Klein-Gordon equation have bounded energy (a necessary physical condition, which anti-commutators do not do).
On the other hand anti-commutators make the Dirac equation (for fermions) have bounded energy (unlike commutators), see spin-statistics connection theorem.
In this sense the anti-commutators is the exact analog of commutators for fermions (but what do actualy commutators mean?). nice and difficult question to answer intuitively.
In a sense commutators (between observables) measure the correlation of the observables. Thus is also a measure (away from) simultaneous diagonalisation of these observables.
Elaborating a little on this.
Lets say we have a state $\psi$ and two observables (operators) $A$, $B$. When these operators are simultaneously diagonalised in a given representation, they act on the state $\psi$ just by a mere multiplication with a real (c-number) number (either $a$, or $b$), an eigenvalue of each operator (i.e $A\psi=a\psi$, $B\psi=b\psi$).
We know that for real numbers $a,b$ this holds $ab-ba=0$ identicaly (or in operator form $(AB-BA)\psi=0$ or $\left[A,B\right]\psi=0$) so the expression $AB-BA=\left[A,B\right]$ (the commutator) becomes a measure away from simultaneous diagonalisation (when the observables commute the commutator is identicaly zero and not-zero in any other case).
Another way to see the commutator expression (which is related to previous paragraph), is as taking an (infinitesimal) path from point (state) $\psi$ to point $A \psi$ and then to point $BA \psi$ and then the path from $\psi$ to $B \psi$ to $AB \psi$. If the operators commute (are simultaneously diagonalisable) the two paths should land on the same final state (point). If not their difference is a measure of correlation (measure away from simultaneous diagonalisation).
When talking about fermions (pauli-exclusion principle, grassman variables $\theta_1 \theta_2 = - \theta_2 \theta_1$),
the commutators have to be adjusted accordingly (change the minus sign), thus become anti-commutators (in order to measure the same quantity).
Continuing the previous line of thought, the expression used was based on the fact that for real numbers (and thus for boson operators) the expression $ab-ba$ is (identicaly) zero.
However fermion (grassman) variables have another algebra ($\theta_1 \theta_2 = - \theta_2 \theta_1 \implies \theta_1 \theta_2 + \theta_2 \theta_1=0$, identicaly).
So what was an identical zero relation for boson operators ($ab-ba$) needs to be adjusted for fermion operators to the identical zero relation $\theta_1 \theta_2 + \theta_2 \theta_1$, thus become an anti-commutator.
Un-correlated observables (either bosons or fermions) commute (or respectively anti-commute) thus are independent and can be measured (diagonalised) simultaneously with arbitrary precision. If not, the observables are correlated, thus the act of fixing one observable, alters the other observable making simultaneous (arbitrary) measurement/manipulation of both impossible.
PS. See how the previous analysis can be generalised to another arbitrary algebra (based on identicaly zero relations), in case in the future another type of particle having another algebra for its eigenvalues appears.
On the other hand anti-commutators make the Dirac equation (for fermions) have bounded energy (unlike commutators)* Do they both have bounded energy?
– MatterGauge Dec 08 '21 at 14:17