In Tent and Ziegler's A Course in Model Theory, they construct a $\phi$-$k$-dividing sequence over any set $A$ of an arbitrary length $\mu$ from a formula $\phi$ with a tree property wrt $k < \omega$ (Lemma 7.2.4.1). The proof is, of course, by transfinite recursion, but they do not state the successor case ("note that we may assume that $\mu$ is a limit ordinal").
In particular, given a sequence of parameters $(a_i \mid i < \mu)$ such that $\{\phi(x, a_i) \mid i < \mu \}$ is consistent and that for each $i < \mu$ $\phi(x, a_i)$ divides over $Aa_{<i}$, I need to find $a_\mu$ to extend the sequence so the new sequence will have the similar property.
Is the idea that I can just take an indiscernible sequence $I$ over $Aa_{<\mu}$ realizing the EM type over $Aa_{<\mu}$ of the parameters witnessing the tree property, and let $a_\mu$ be the first element of $I$?
"We may assume that $\mu$ is a limit ordinal" just means that if you want to build a $\phi$-$k$-dividing sequence of length $\mu$, you might as well build one of length $\mu + \omega$ (a limit ordinal) and then take the initial subsequence of length $\mu$.
There's no reason to take indiscernible sequences in this proof. Remember that "$\phi(x,a)$ divides with respect to $k$ over $A$" just means that there is an infinite sequence $(a_i)_{i\in\omega}$ of realizations of $\text{tp}(a/A)$ such that $\{\phi(x,a_i)\mid i\in\omega\}$ is $k$-inconsistent.
The crucial step in the proof is to apply compactness to get a tree of height $\mu$ which is $\kappa$-branching, where $\kappa$ is larger than the number of types over a set of size $\max(|A|,|\mu|)$. Let's say we've found a partial path through the tree $(a_\alpha)_{\alpha\leq\lambda}$ for $\lambda < \mu$ which is a $\phi$-$k$-dividing sequence. Then since there are $\kappa$-many children of $a_\lambda$, infinitely many have the same type over $A(a_\alpha)_{\alpha\leq\lambda}$, so we can choose one of them to be $a_{\lambda+1}$, and $\phi(x,a_{\lambda+1})$ divides with respect to $k$ over $A(a_\alpha)_{\alpha\leq\lambda}$.