Say we have a branched $G$-Galois covering $X \rightarrow Y$ of surfaces, and assume that the branch locus (in $Y$) is a divisor with normal crossings. I will always assume that the ramification is tame, and you can pretend we're doing this with varieties over $\mathbb{C}$ if you prefer. This implies that the inertia groups of any two irreducible components of the ramification divisor commute, because they live in the inertia group of the point which is abelian. I've been told the following fact, and have been spending way too much time trying to understand why it's true:

Say the irreducible components $D_1$ and $D_2$ of the branch locus meet at a node. Let's say we pick a $P_1$ and $P_2$ irreducible preimages of $D_1$ and $D_2$ respectively, that meet.

If we blow up that node of the branch locus in $Y$ (and get the new scheme/variety $Y'$) and then normalize in $\kappa(X)$ (to get $X'$), we get a map $X' \rightarrow Y'$ where over $D_1$, $D_2$ and $E$ (the exceptional divisor) are $P_1$, $P_2$ and an irreducible divisor connecting them $P_3$; such that if $I_1$ was the inertia group of $P_1$ and $I_2$ the inertia group of $P_2$ then $I_1I_2$ (which is defined because $I_1$ and $I_2$ commute) is the inertia group of $P_3$.

Why is this true? I've been blowing things up and normalizing for way too long and it has been thoroughly uninsightful. I must be missing an important heuristic.