Pullback by a pointwise product Let $M$ be a compact manifold without boundary and let $f, g : M \to S^1$ be smooth maps into the unit circle. Then $\phi := fg : M \to S^1$ is still smooth and we can pullback the standard volume (1-)form ${\rm d} \theta$ of $S^1$ by $\phi$, i.e. evaluate $\phi^*({\rm d}\theta)$. Of course, by definition of pullback and setting $\omega := {\rm d}\theta$ for convenience, we have
$$\forall p \in M, \forall v \in T_p M, \quad (\phi^*\omega)_p(v) = \omega_{\phi(p)} ({\rm d}\phi_p(v))$$
Here, ${\rm d}\phi_p : T_p M \to T_{\phi(p)} S^1$ is the differential of $\phi$ at $p$. Now, I would like to write the above formula in terms of $f$ and $g$. Clearly,
$$\forall p \in M, \forall v \in T_p M, \quad ((fg)^*\omega)_p(v) = \omega_{(fg)(p)} ({\rm d}(fg)_p(v)).$$
If we look at $S^1$ as the unit circle into $\mathbb{C}$, we can write $$\forall p \in M, \quad {\rm d}(fg)_p = f(p) {\rm d} g_p + g(p) {\rm d} f_p.$$ Since $f(p)$ and $g(p)$ can be regarded as isometries of $S^1$, both  $f(p) {\rm d} g_p$ and $g(p) {\rm d} f_p$ can be viewed as maps from $T_p M$ into $T_{(fg)(p)} S^1$, and this holds for each $p \in M$, so that the computation seems to me to be consistent up to now. Notice that, of course, the order of $f$ and $g$ is inessential in the product $fg$ and hence in the previous computations. Now, by linearity we have
$$\omega_{f(p)g(p)}({\rm d}(fg)_p(v)) = \omega_{f(p)g(p)}(f(p){\rm d}g_p(v)) + \omega_{f(p)g(p)}(g(p){\rm d}f_p(v)),$$
for all $p \in M$ and $v \in T_p M$. At this point, I would like to write the two terms on the right in terms of pullbacks by $f$ and $g$ but I am not really sure how to go on and conclude in a clean way. I have no clues from standard textbooks, as this situation is not usually handled in them (to the best of my knowledge, at least), because in general the (pointwise) product of two maps $f,g:M \to N$ does not make sense or at least does not take values into $N$.
My interest in this computation comes from the special case in which $f$ is a harmonic map and $g$ is of the form $e^{i\psi}$ for some smooth function $\psi : M \to \mathbb{R}$, in which (reading some papers) I would expect to get something like $\phi^* \omega = f^*\omega + {\rm d}\psi$. Notice, if useful, that $f^*\omega$ is a harmonic 1-form since $f$ is harmonic. So, how to conclude the computation above, and how to get the mentioned special result?
 A: The expression should be $(fg)^*\omega=f^*\omega+g^*\omega$. Since you regard the circle group $S^1$ as a subset of $\mathbb{C}$, we can think of the one-form $\omega=d\theta$ as a form which maps complex numbers tangent to the circle to real numbers. One choice for this is
$$
\omega_{u}(z)=\frac{z}{iu}=-i\bar{u}z
$$
where $u\in S^1\subset\mathbb{C}$ and $z\in\mathbb{C}$ is orthogonal to $u$. Showing the equality from there is a matter of computation.
Alternately, if you're willing to add more abstraction, there's a much more general approach using Lie group-valued functions and Lie algebra-valued forms. Given a map $f:M\to G$ between asmooth  manifold $M$ and a Lie group $G$ with Lie algebra $\mathfrak{g}$, we can regard the derivative of $f$ as a $\mathfrak{g}$-valued one form given by $f^*\omega$ where $\omega$ is the Maurer-Cartan form of $G$. With a bit of work, one can show that the pointwise product obeys the following product rule:
$$
(fg)^*\omega=\operatorname{Ad}_{g^{-1}}(f^*\omega)+g^*\omega
$$
Here $\operatorname{Ad}$ is the adjoint representation. Your question is the special case of $G=S^1$; the adjoint doens't appear since the group is Abelian.
