Basic definitions Let $X$ be a smooth variety and $Z\subset X$ a hypersurface. Let $f$ be a local equation for $Z$ near $p\in X$, then we say that the multiplicity of $Z$ at $p$ is the largest $k$ such that $f\in \mathfrak{m}^k_p$ where $\mathfrak{m}_p$ is the maximal ideal of the local ring $\mathcal{O}_{X,p}$. We write this multiplicity as $\text{mult}_p(Z)$.
Set up Let $\mathbb{P}^d = \mathbb{P}H^0(\mathcal{O}_{\mathbb{P}^1}(d))$ be the projective space of degree $d$ divisors on $\mathbb{P}^1$. Note that we have a hypersurface $\mathcal{D} \subset \mathbb{P}^d$ , which we'll call the discriminant hypersurface, defined as $\{ F \in \mathbb{P}^d : \text{$F$ has at least one double root}\}$. So $\mathcal{D}$ parametrizes "singular" degree $d$ divisors on $\mathbb{P}^1$.
Question Is the following claim true? Let $F\in \mathbb{P}^d$, then $\text{mult}_F(\mathcal{D}) = \sum_{p\in\mathbb{P}^1 : F(p) = 0} (\text{ord}_p(F) - 1)$.
Recall that here $\text{ord}_p(F)$ means the order of vanishing of $F$ at $p$: the largest $k$ such that $F\in \mathfrak{m}^k_p$ in $\mathcal{O}_{\mathbb{P}^1,p}$.
I have some suspicion to think the above claim is true. For example, I know that a polynomial $F\in\mathbb{P}^d$ is a smooth point of $\mathcal{D}$ if $F$ has only one double root (meaning the order is $2$) and all other roots are simple roots (meaning the order is $1$), and in this case our formula gives us $\text{mult}_F(\mathcal{D}) = 1$; which precisely means that $\mathcal{D}$ is smooth at $F$.
I tried to prove this formula, but have so far failed. One idea I had was to use the the space $X = \{(F,p)\in\mathbb{P}^d\times\mathbb{P}^1 : F(p) = 0\}$, which contains the hypersurface $Y = \{(F,p)\in X : \text{ord}_p(F)\geq 2\}$. We then have the projection $\pi: X\rightarrow \mathbb{P}^d$, and I think here we have $\pi^\ast\mathcal{I}_{\mathcal{D}/\mathbb{P}^d}= \mathcal{I}_{Y/X}$. Note that for $F\in \mathcal{D}$ the fiber $\pi^{-1}(F) = \{p \in \mathbb{P}^1 : F(p) = 0\}$ contains the locus $\{p\in\mathbb{P}^1 : \text{ord}_p(F)\geq 2\}$, which is $Y\cap \pi^{-1}(F)$. I wasn't quite sure how to get this to work, but I think conjectured that
1 $\text{mult}_{(F,p)}(Y) = \text{ord}_p(F) - 1$
2 $\text{mult}_{F}(\mathcal{D}) = \sum_{p\in \pi^{-1}(F)} \text{mult}_{(F,p)}(Y)$
Putting together 1 and 2 would then answer the above question affirmatively. I'd really appreciate any help on this. Thanks!