I am not sure what kind of answer you are looking for. But if you have a stack $X$, then the inertia stack $IX$ is basically the gadget parametrizing pairs $(x,\sigma)$ where $x$ is a point of $X$ and $\sigma$ is in the isotropy group at $x$ (an automorphism of $x$). Informally, the locus where you have automorphism group $G$ becomes "doubled" $|G|$ times. Example: for the stack $[\mathbb A^1/\mu_2]$ the inertia stack is $[\mathbb A^1/\mu_2] \sqcup B\mu_2$, the extra $B\mu_2$ corresponding to the origin being doubled because it has an extra automorphism.
One way to think about it is as a kind of "infinitesimal loop space", where instead of taking maps to $X$ from a circle we take maps from the homotopically equivalent object $B\mathbb Z$. This is pleasant because the inertia stack is the fibered product $X \times_{X\times X} X$, and for a topological space $X$ the homotopy fibered product $X \times_{X\times X}^h X$ is the space of free loops on $X$.
You can motivate the inertia stack through Gromov--Witten theory. If $X$ is a variety, then there is an evaluation map from the stack of stable $n$-pointed maps to $X$ to $X^n$. If $X$ is a stack, then the correct notion is that of a twisted stable map, and in this case the evaluation maps do not land on $X$ but in its inertia stack $IX$! (In fact it lands on the rigidified inertia stack, where some automorphisms have been removed from the picture, but nevermind this). So quantum cohomology of a stack is not extra structure on the cohomology ring of $X$ itself, but on the cohomology ring of $IX$.