Possible inconsistency of weakly Shelah cardinals (I hope not)

Question

A Mathoverflow question by Trevor Wilson defines weakly Shelah cardinals as follows:

A cardinal $\kappa$ is weakly Shelah if for all $f : \kappa \to \kappa$ there is some $\alpha < \kappa$ that is closed under $f$ and there is some elementary embedding $j : V \to M$ (where $M$ is a transitive class) such that $\operatorname{crit}(j) = \alpha$ and $j(\alpha) > \kappa$ and $V_{j(f)(\kappa)} \subset M$.

Edit: Because of the issue noted by Sean Cox in the comments, a better definition also requires that $j(f)(\beta)=f(\beta)$ whenever $\beta \lt \kappa$.

If $f$ is the function enumerating ordinals $\beta$ such that $V_\beta \prec V_\kappa$ (and there exists $\kappa$-many such ordinals since $\kappa$ is inaccessible so $V_\kappa \vDash KM$ and the argument in the blog post Kelley–Morse set theory implies Con(ZFC) and much more by Joel David Hamkins applies) but are not limits of such ordinals, then $V_{\beta_1} \prec V_{\beta_2}$ whenever $\beta_1$ and $\beta_2$ are in the range of $f$ or limits of ordinals in the range of $f$. By elementarity of $j$, $V_\kappa \prec V_{j(f)(\kappa)}$. Thus $\kappa$ is 0-extendible (defined by Bagaria et al. 2015, Superstrong and other large cardinals are never Laver indestructible, also known as otherworldly).

However, a comment by Trevor Wilson says that every $\Sigma_3$-reflecting Woodin cardinal is weakly Shelah:

On further thought, it seems like $\kappa$ Woodin and $\Sigma_3$-reflecting implies κ weakly Shelah: given $f:\kappa \to \kappa$, use Woodinness to get some $\alpha \lt \kappa$ that is $\lt \kappa$-$f$-strong. Then the desired conclusion (on existence of $j$) holds for cofinally many $\bar\kappa \lt \kappa$ in place of $\kappa$. Formulating this in terms of extenders, $\Sigma_3$-reflection implies the desired conclusion holds for cofinally many $\bar\kappa \in Ord$ in place of $\kappa$, and therefore for $\kappa$ itself (since $f$ is increasing.)

If $V_\kappa \prec V_{j(f)(\kappa)}$ and $\kappa$ is Woodin, then $V_{j(f)(\kappa)} \vDash \text{"$\kappa$ is $\Sigma_3$-reflecting and Woodin"}$. That would mean that there cannot be a least $\lambda$ such that $V_\lambda \vDash \text{"$\kappa$ is weakly Shelah"}$ and the existence of $\Sigma_3$-reflecting Woodin cardinal is inconsistent. Where does my argument or Trevor Wilson's go wrong?

Answer 1

This question was answered in the comments:

As Gabe Goldberg pointed out, Trevor Wilson's argument that $\Sigma_3$-reflecting Woodin cardinals are weakly Shelah fails because the reflection argument would have to use the function $f$ as a parameter, and $\kappa$ being $\Sigma_3$-reflecting is not enough for that.

Additionally, as Sean Cox pointed out, we can't be sure that $V_\kappa \prec V_{j(f)(\kappa)}$ since it is possible that there is some $\beta \lt \kappa$ such that $j(f)(\beta) \gt \kappa$. I assumed that $j(f) \upharpoonleft \kappa = f$ but I don't know how to prove that from the definition.