Here is a fairly straightforward proof which also proves various generalizations of your problem. Choose $c,d$ such that $\phi_2(c,d) \neq 0$. If no such $c,d$ exist, then $f$ is identically $0$ and can be completely factored trivially. Now,
$$\phi_1(x_1, x_2)=\psi_1(x_1, c)\psi_2(x_2, d) \phi_2(c,d)^{-1},$$
for all $x_1,x_2$. Similarly, choosing $a,b$ such that $\phi_1(a,b) \neq 0$, we have
$$\phi_2(x_3, x_4)=\psi_1(a, x_3)\psi_2(b, x_4) \phi_1(a,b)^{-1},$$
for all $x_3,x_4$. Thus,
$$f(x_1 ,x_2 ,x_3 ,x_4)=\phi_1(a,b)^{-1}\phi_2(c,d)^{-1}\psi_1(x_1, c)\psi_2(x_2, d) \psi_1(a, x_3)\psi_2(b, x_4), $$
for all $x_1,x_2,x_3,x_4$. $\Box$

The same proof also proves the following generalization. Given a partition $\alpha$ of $[n]$, we say that $f(x_1, \dots, x_n)$ *factors with respect to $\alpha$* if for each $A \in \alpha$ there exists a function $f_A$ (which only depends on the variables $x_i$ for $i \in A$) such that $f(x_1, \dots, x_n)=\prod_{A \in \alpha} f_A$. Given two partitions $\alpha$ and $\beta$ of $[n]$, $a \wedge b$ is the partition of $[n]$ whose sets are the non-empty sets of the form $A \cap B$ for $A \in \alpha$ and $B \in \beta$.

**Lemma.** Let $\alpha$ and $\beta$ be partitions of $[n]$. If $f(x_1, \dots, x_n)$ factors with respect to both $\alpha$ and $\beta$, then $f(x_1, \dots, x_n)$ factors with respect to $\alpha \wedge \beta$.

Note that I am only using the fact that the function takes values in some field or some group. I am not sure if the result still holds if inverses do not exist (this was asked by Richard Stanley in the comments below).