Here is something very close which might be adaptable to get a suitable $f$ and ${\leq_f}$. I apologize for the relative complexity of the argument but I couldn't find any way to simplify it...

Let $\phi_0,\phi_1,\ldots$ be a computable enumeration of all sentences in the language of set theory. I will assume that $\phi_0$ is your favorite contradiction, which I will also denote $\perp$.

I will define a function $f:\omega\to\omega$ and computable linear ordering ${\preceq}$ of $\omega$ by stages. At each stage $s$, I will decide the restrictions of $f$ and ${\preceq}$ on the set $\{0,1,\dots,t_s\}$. I will also keep track of a marker $m_s$ which will mark the ${\preceq}$-first element of $\{0,1,\dots,t_s\}$ for which it is "known by stage $s$" that $ZFC + \{\phi_{f(i)} : i \leq t_s \land i \preceq m_s\}$ is inconsistent. By the phrase "known by stage $s$" I mean that $\perp$ appears by stage $s$ in the "standard computable enumeration" of the consequences of $ZFC + \{\phi_{f(i)} : i \leq t_s \land i \preceq m_s\}$. This standard computable enumeration is only allowed to enumerate finitely many consequences at each stage. I will also assume that this standard computable enumeration is such that:

1. If $F$ is a finite set of sentences in the language of set theory, then all elements of $F$ appear at stage $0$ of the enumeration of the consequences of $ZFC + F$.

2. If $F \subseteq G$ are finite sets of sentences in the language of set theory, then all elements which appear by stage $s$ of the enumeration of the consequences of $ZFC + F$ also appear by stage $s$ of the enumeration of the consequences of $ZFC + G$.

Stage $s = 0$ is trivial: $t_s = 0$, $f(0) = 0$, $0 \preceq 0$. (Note that $m_0 = 0$ since $\phi_0 = \perp$.)

At stage $s+1$, let $\ell_s$ be the first element of the set $\{i \leq t_s : i \preceq m_s\}$ (in the usual ordering) such that $ZFC + \{\phi_{f(i)} : i \leq t_s \land i \preceq \ell_s\}$ is known by stage $s+1$ to be inconsistent. Note that $\ell_s$ necessarily exists since $m_s$ is in the set being searched.

Case $\ell_s \neq m_s$: Then simply let $t_{s+1} = t_s$. (Note that $m_{s+1} = \ell_s$.)

Case $\ell_s = m_s$: Then consider the first element $n$ of the set
$$\{0,1,\ldots,s+1\}\setminus\{f(i): i \leq t_s \land i \prec m_s\}$$
such that
$$ZFC + \phi_n + \{\phi_{f(i)} : i \leq t_s \land f(i) < n \land i \prec m_s\}$$
is not known to be inconsistent by stage $s+1$.

* If there is no such $n$, then simply let $t_{s+1} = t_s$. (Note that $m_{s+1} = m_s$.)
* Otherwise, set $t_{s+1} = t_s + 1$ and $f(t_{s+1}) = n$. Place $t_{s+1}$ immediately after the maximum element of $\{i \leq t_s : f(i) < n \land i \lneq_f m_s\}$ in the ${\preceq}$-ordering.

Note that we always have that if $i \prec j \prec m_s$ and $i, j \leq t_s$ then $f(i) < f(j)$.

It is straightforward to check that $\sup_{s<\omega} t_s = \omega$. Thus $f:\omega\to\omega$ is well-defined and ${\preceq}$ is a linear ordering of $\omega$.

Observe that every $n \in \omega$ has either finitely many ${\preceq}$-predecessors or finitely many ${\preceq}$-successors. The dividing line is whether the theory $ZFC + \{\phi_{f(i)} : i \preceq n\}$ is consistent or not.

*Proof sketch.* First suppose that $ZFC + \{\phi_{f(i)} : i \preceq n\}$ is inconsistent. Then, there is some stage $s$ such that $n \leq t_s$ and
$$ZFC + \{\phi_{f(i)} : i \leq t_s \land i \preceq n\}$$
is inconsistent and this is known by stage $s$. It follows that $m_s \preceq n$ and after this stage no new elements will appear after $n$ in the ${\preceq}$-ordering after stage $s$.

If $ZFC + \{\phi_{f(i)} : i \preceq n\}$ is consistent. Then note that $f(i) < f(j) < f(n)$ for all $i \prec j \prec n$ and so $n$ has no more than $f(n)$ predecessors in the ${\preceq}$-ordering. *QED*

Now let $I$ be the set of all $n \in \omega$ which have only finitely many ${\preceq}$-predecessors. I claim that $ZFC + \{\phi_{f(i)}:i \in I\}$ is complete.

*Proof sketch.* Show by induction that for each $n$, there is a stage $s \geq n$ such that either $\phi_n \in \{\phi_{f(i)} : i \leq t_s \land i \in I\}$, or $ZFC + \phi_n + \{\phi_{f(i)} : i \leq t_s \land i \in I\}$ is known to be inconsistent by stage $s$.

Suppose we know the desired fact for all $m < n$. If $ZFC + \phi_n + \{\phi_{f(i)} : i \in I\}$ is inconsistent, then simply find a stage $s$ such that $t_s$ is large enough to witness this. Otherwise, find a stage $s_0 \geq m$ large enough to witness the desired fact simultaneously for all $m < n$. Then, find a stage $s_1 \geq s_0$ such that $\{ i \leq t_{s_0} : i \in I \} = \{i \leq t_{s_0} : i \prec m_{s_1} \}$. Note that for every stage $s \geq s_1$, we then have that
$$ZFC + \phi_n + \{\phi_{f(i)} : i \leq t_s \land f(i) < n \land i \prec m_s\}$$ 
is consistent (hence not known to be inconsistent by stage $s$).
Furthermore, if $n \notin \{f(i) : i \leq t_s \land i \prec m_s\}$ then $n$ is necessarily the first element of the set
$$\{0,1,\ldots,s+1\} \setminus \{f(i) : i \leq t_s \land i \prec m_s\}$$
such that 
$$ZFC + \phi_n + \{\phi_{f(i)} : i \leq t_s \land f(i) < n \land i \prec m_s\}$$ 
is not known to be inconsistent by stage $s+1$. Waiting a few more stages if necessary, we will eventually have $\ell_s = m_s$ and then $n = f(t_s+1)$. Moreover, the hypotheses on $n$ guarantee that we will then have $t_s+1 \in I$. *QED*