It is true for $k=1,2$ but not for $k\ge 3$.
Write $f=g^2$ near $0$ and let $g(z)=\sum_{m\ge 0}a_m z^m$. Then $|f|(z)=\overline{g(z)}g(z)$, so we want (up to $k!$ on both sides) that
$$
\left|\sum_{0\le m\le k/2}a_ma_{k-m}\right|\le \max_u\left|\Re\sum_{0\le m\le k/2}\bar a_ma_{k-m}P_{m}(u)\right|
$$
where $P_{m}(u)$ is the average of all products of $u_1,\dots,u_k$ with exactly $m$ conjugation bars over $u$'s and if $k=2m$, then the last term in the sum should be taken with the coefficient $1/2$.
Now notice that if we multiply $a_m$ and $a_{k-m}$ by any $\zeta_m\in\mathbb T$, then we can rotate the terms on the LHS in any way we want without changing the ones on the RHS. Also we can make $\bar a_ma_{k-m}$ any numbers we want except $a_m\bar a_m$ should be positive when $k$ is even and $m=k/2$.
Thus, the desired inequality is equivalent to
$$
\sum_{0\le m\le k/2}|b_m|\le \max\left|\Re\sum_{0\le m\le k/2}b_mP_m(u)\right|
$$
However, the only way to achieve this is to align all terms in all $b_mP_m(u)$ with either $+1$ or $-1$ simultaneously.
If $k=1$, then $P_0(u)=u_1$, so it is not a problem.
If $k=2$, then $P_0(u)=u_1u_2$ and $P_1(u)=\frac 12(u_1\bar u_2+u_2\bar u_1)$ and we know that $b_1$ is already aligned with $+1$, so we can align $P_1$ with $+1$ by choosing $u_1=u_2=\zeta\in\mathbb T$ and we still have enough freedom to rotate $u_1u_2$ to anything we want.
However, when $k>2$, we have arbitrary $b_0$ and $b_1$ and aligning all terms in $P_1$ between themselves requires that all $u_k=\pm\zeta$ for some $\zeta\in \mathbb T$. Thus, our sum starts with $\pm (b_0\zeta^k+b_1\zeta^{k-2})$ and for generic $b_0,b_1$ we just don't have enough freedom to bring even these two terms to the real line simultaneously.
