Prescribed-start and starting-point-free serial choice are equivalent in ZF

Assume prescribed-start DC. Given nonempty $A$ and serial $R$, fix one $a_0\in A$. Its prescribed chain is a chain with unrestricted start, so starting-point-free DC follows. This is a single existential instantiation, not a family of selections.

Conversely assume starting-point-free DC, and fix nonempty $A$, serial $R$ and $a_0\in A$. By Separation in $\omega \times \mathcal{P}(\omega \times A)$, the pairs $(l,p)$ with $l\ge 1$, $p:l\to A$, $p(0)=a_0$, and $p(i)Rp(i+1)$ for every $i+1<l$ form a set $P$. The pair $(1,\{(0,a_0)\})$ belongs to $P$, since there are no adjacent coordinates to check.

Define $S$ on $P$ by $(l,p)S(l+1,q)$ exactly when $q$ extends $p$. This is a subset of $P\times P$. For any $(l,p)\in P$, seriality gives one $b\in A$ with $p(l-1)Rb$; the function $q=p\cup \{(l,b)\}$ has domain $l+1$ and satisfies all required edges, old ones from $p$ and the new last edge by the choice of $b$. Thus $S$ is serial. No simultaneous successor function has been selected.

Apply starting-point-free DC to the nonempty set $P$ and serial $S$. It gives $h(k)=(l_k,p_k)$ with $p_{k+1}$ extending $p_k$ and $l_{k+1}=l_k+1$. Induction gives $l_k=l_0+k$ and, for $j\le k$, $p_j\subseteq p_k$: the zero case is reflexivity, and each successor uses one end extension. In particular the domains are unbounded in $\omega$.

Replacement gives the set $\{p_k:k\in \omega \}$ and Union gives $f={\bigcup }_{k\in \omega }{p}_k$. Two pairs in $f$ with the same first coordinate lie together in $p_{\max (j,k)}$, so have the same second coordinate. Every $n$ lies in $l_{n+1}$, since $l_0\ge 1$, and all domains lie in $\omega$. Consequently $f:\omega \to A$ and $f(0)=a_0$.

For any $n\in \omega$, both $n$ and $n+1$ lie in $l_{n+2}$. That path's edge gives $f(n)=p_{n+2}(n)R{p}_{n+2}(n+1)=f(n+1)$. Hence $f$ is the prescribed chain. Together with the first implication this proves the equivalence, without any additional choice axiom.