Let $A$ be a local ring, which we can assume is reduced. Let $k$ be the residue field of $A$.

In the Stacks project (https://stacks.math.columbia.edu/tag/06DT), I have learned some notion of the number of "geometric branches" of $A$ as being the number of minimal primes of the strict henselianization of $A$. Equivalently (as shown in that Stacks project page), it is the number of maximal ideals $\mathfrak{m}'$ of $A'$ ($A'$ being the integral closure of $A$ in its total ring of fractions), each weighted by the separable degree of $A'/\mathfrak{m}'$ over $k$.

This definition makes sense to me intuitively, on the one hand because if $A$ is the stalk of the structure sheaf of some curve $X$ at a point $x \in X$, then the etale topology on $X$ (the structure sheaf on which $A^{sh}$ is the stalk of) should be fine enough to distinguish between the branches of $X$ passing through $x$, and because on the other hand taking the normalization of a curve at a point (similarly to a blowup) is supposed to separate the branches.

However, I am having difficulty working out all of this in the simplest possible geometric example. For simplicity, let $k$ be an algebraically closed field. As we know from calculus or undergraduate algebraic geometry, if we have a (affine since we only care about the local ring) curve $X$ in $\mathbf{A}_k^2$ cut out by a polynomial $$f = \sum_{i \geq 1} f_i \in k[X, Y],$$ $f_i$ being homogeneous of degree $i$ (the absence of constant term meaning we are assuming the curve passes through the origin), then the tangent lines to $X$ at the origin are cut out by $f_{i_0}$, where $i_0 \geq 1$ is the smallest $i$ such that $f_i \neq 0$. So we should expect that $\mathcal{O}_{X, (0, 0)}$ has at most $i_0$ branches according to the above definition, with equality if and only if $f_{i_0}$ factors over $k$ into $i_0$ distinct linear factors. Is this true ?

I can make some sense of this in one special case: if $i_0 = 2$ and $f_{i_0}$ has no repeated factors, then WLOG we can assume $f_{i_0} = (X-Y)(X+Y)$, and I can show that $\mathcal{O}_{X, (0, 0)}^{sh}$ is isomorphic to the strict henselianization to the localization at $(X, Y)$ of $k[X, Y]/(XY)$, because if we send $X$ to $\alpha X + \beta Y$ and $Y$ to $\alpha X - \beta Y$ then $XY$ maps to $\alpha^2X^2 - \beta^2Y^2$, and since $\mathcal{O}_{X, (0, 0)}^{sh}$ is henselian we can choose $\alpha, \beta \in 1 + \mathfrak{m}$ to be such that this is exactly $f$, and it is easy to check (thanks to where $\alpha, \beta$ live) that this will provide an isomorphism of henselian local rings. This means the number of branches is the same as if we had no terms after $f_{i_0}$, i.e. if $A = (k[X, Y]/(XY))_{(X, Y)}$, which is explicit enough that we can compute the number of branches to be $2$ (for example by using the equivalent definition involving the normalization).

But in the case where there are repeated linear factors, or more than 2 linear factors, I am completely stuck (in particular from examples it no longer seems to be true that the henselianization is indifferent to the terms after $f_{i_0}$). Is my claim about the branches still true ? Am I just missing some algebraic manipulation with the henselianization, or is there a conceptual step that I have not figured out ?