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 $f(x_1, \dotsc, x_n)$ 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 $f(x_1, \dotsc, x_n)=\prod_{A \in \alpha} f_A$. 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 $f(x_1, \dotsc, x_n)$ factors with respect to both $\alpha$ and $\beta$, then $f(x_1, \dotsc, x_n)$ 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).
Update. The above lemma does not always hold for monoids, as shown by Harry West in an answer to Functions over monoids which factor in two different ways.
