Yes, equivalence can be checked in polynomial time (in fact, in quadratic time).
I will describe how to test whether $\psi_1 \lor \neg \psi_2$ is true for all assignments. You can do the same for $\neg \psi_1 \lor \psi_2$, and use this to test whether $F$ is a tautology, i.e., whether $\psi_1,\psi_2$ are logically equivalent.
I will do this by checking whether $\psi_1 \lor \neg \psi_2$ is false for any assignment, or in other words, whether $\neg(\psi_1 \lor \neg \psi_2)$ is satisfiable. Notice that
$$\neg(\psi_1 \lor \neg \psi_2) = \neg \psi_1 \land \psi_2,$$
so it suffices to test satisfiability of $\neg \psi_1 \land \psi_2$ where $\psi_1,\psi_2$ are Krom (2-CNF) formulas.
Suppose that $\psi_1 = c_1 \land \cdots \land c_k$ where $c_i$ is the $i$th clause in $\psi_1$. Then
$$\neg \psi_1 = (\neg c_1) \lor \cdots \lor (\neg c_k).$$
\neg \psi_1 \land \psi_2 &= ((\neg c_1) \lor \cdots \lor (\neg c_k)) \land \psi_2\\
&= (\neg c_1 \land \psi_2) \lor \cdots \lor (\neg c_k \land \psi_2).
Now, $\neg \psi_1 \land \psi_2$ is satisfiable iff $\neg c_i \land \psi_2$ is satisfiable for some $i$. So, we can iterate over $i$ and test satisfiability of each $\neg c_i \land \psi_2$; if any of them are satisfiable, then $\neg \psi_1 \lor \psi_2$ is satisfiable and $F$ is not a tautology and $\psi_1,\psi_2$ are not logically equivalent.
How to test satisfiability of $\neg c_i \land \psi_2$? Well, $c_i$ has the form $(\ell_1 \lor \ell_2)$ where $\ell_1,\ell_2$ are two literals, so $\neg c_i \land \psi_2$ has the form $\neg \ell_1 \land \neg \ell_2 \land \psi_2$. This is also a Krom (2-CNF) formula, so you can test its satisfiability using the standard polynomial-time algorithm. You do a linear number of such tests, so the total running time is polynomial. In fact, it is quadratic, as testing satisfiability can be done in linear time.