Why is weak Kőnig's lemma weaker than Kőnig's lemma?

The issue is that for a finitely branching subtree $T$ of $\omega^{<\omega}$, the function $f$ mapping $\sigma$ to the greatest $n$ such that the concatenation $\sigma ^\frown n$ is in $T$ may not be computably bounded.

So $f$ may not "exist" in your model. (Even though the model knows that for each $\sigma$ such an $n$ exists.)

However $ACA_0$ is enough I believe, because you can repeatedly ask a Halting Problem oracle

"is there an $m>k$ with $\sigma ^\frown m\in T$?"

until the answer comes back as "no".

Interestingly, even Kőnig's lemma restricted to subtrees of $\omega^\omega$ for which every node has at most two children is equivalent to the full Kőnig's lemma, not weak Kőnig's lemma. This is because you can make such a tree $T$ for which an infinite branch gives you a halting oracle: For each $\sigma\in T$ of length $n$, put $\sigma\frown0$ in $T$, and if the $n$th program halts in $k$ steps, also put $\sigma\frown k$ in $T$, and cut off all descendents of $\sigma\frown0$ of length $n+k$ (that is, instead of following the previous rules for creating child nodes, don't create any child nodes). This is a computable tree, but the only infinite branch has a $0$ in the $n$th place iff the $n$th program does not halt. This can be relativized to any oracle, and hence gives you $ACA_0$, which as Bjorn Kjos-Hanssen noted, is enough to prove Kőnig's lemma.

And Kőnig's lemma for subtrees of $n^\omega$ is equivalent to weak Kőnig's lemma. The proof of Kőnig's lemma from weak Kőnig's lemma that you described can be carried out in $RCA_0$ in this case. Same goes if you let nodes in deeper levels have more children, just so long as there is a computable bound for the labels of the children that any given node can have.

This is all just to emphasize what Bjorn already hinted at: binary vs finitely-branching is sort of a red herring; the difference is whether or not there is a function bounding how far you have to look for child nodes.