The theorem of transfinite recursion states the following (A quick introduction to basic set Theory).
Theorem 4.4 (Transfinite recursion). For every ordinal $\kappa$, set $A$, and map $^3 F: \operatorname{Partial}(\kappa, A) \times \kappa \rightarrow A$, there is a unique function $f: \kappa \rightarrow A$ such that for every $\alpha \in \kappa$,
$$
f(\alpha)=F\left(\left.f\right|_\alpha, \alpha\right) .
$$
The proof of this just uses transfinite induction. If we just take $\kappa = \omega$ (the first limit ordinal), and $F(g, n) = f(n)$ where $g:\{1,\dots, n-1 \} \to A$, it sounds really similar to the dependent axiom of choice. I guess the only part in the dependent axiom of choice that is different is that $g$ should be $\{ n-1 \}\to A$, i.e. it only considers the immediate predecessor, but I don't think it will make much of a difference.
So why doesn't transfinite recursion imply the axiom of dependent choice?
Edit: I just realized now, but in DC, we can have aRb and aRc, but in this case, the next step is determined by the function F, so maybe that's why?
Proof. We prove by transfinite induction on $S(\kappa)$ that for all $\gamma \leqslant \kappa$, there is a unique function $f_\gamma: \gamma \rightarrow A$ satisfying
$$
f_\gamma(\alpha)=F\left(\left.f_\gamma\right|_\alpha, \alpha\right),
$$
for every $\alpha<\gamma$. Granted this, we would take $f=f_\kappa$ and be done.
For $\gamma=0$, put $f_\gamma=\emptyset$ and note that this is the only function defined on $\emptyset$ in general. For $\gamma$ a limit ordinal, it follows from uniqueness that for all $\alpha<\beta<\gamma, f_\alpha \subseteq f_\beta$ (see Remark 2.12). Thus $f_\gamma=\bigcup_{\beta<\gamma} f_\beta$ is a function from $\gamma$ to $A$ and it clearly satisfies $(*)$ as so do the functions $f_\beta, \beta<\gamma$. The uniqueness of $f_\gamma$ also follows from the uniqueness of $f_\beta$ for each $\beta<\gamma$.
For the successor case, let $\gamma=S(\beta)$. For $\alpha \in \gamma$, put
$$
f_\gamma(\alpha)=\left\{\begin{array}{ll}
F\left(f_\beta, \beta\right) & \text { if } \alpha=\beta \\
f_\beta(\alpha) & \text { otherwise }
\end{array} .\right.
$$
It is clear that $f_\gamma$ satisfies $(*)$ and that it is the unique such function.
Best Answer
Recursion, in all of its forms, require you to have an explicit step function. Namely, there is a function that tells you exactly how to proceed from a given situation.
We, the mathematicians, who often work under assumptions of choice, often forego that explicit function in favour of merely proving that a candidate for that next step exists. What we're hiding is that the step function is really a choice function from these sets of candidates. Of course, we know that sometimes this approach is necessary.
Indeed, if you allow to redefine recursion in that "sloppy" form, you will get back some dependant choice, or even the full axiom of choice depending on how exactly you've phrased it.