Let me give this a shot. I'm bound to mess up signs, so assume $p = 2$. Write $\langle \cdot, \cdot \rangle$ for the pairing between $\mathcal{A}_*$ and $\mathcal{A}^*$.
Lemma. $\newcommand{\Sq}{\mathrm{Sq}}\langle \chi \xi_k, \Sq^r \rangle = \delta_{r = 2^k-1}$.
Proof. We have
\begin{align}
\langle \chi \xi_k, \Sq^r \rangle &= \langle \xi_1^{2^k-1} \bmod{(\xi_2, \xi_3, \ldots)}, \Sq^r \rangle \\
&= \langle \xi_1^{2^k-1}, \Sq^r \rangle \\
&= \delta_{r = 2^k-1}.
\end{align}
Proposition. $\Sq^r_* \chi \xi_i = \begin{cases} \chi \xi_{i-k}^{2^k}, & \text{if } r = 2^k - 1 \\ 0, & \text{otherwise}. \end{cases}$
Proof. For any $x \in \mathcal{A}^*$, we have
\begin{align}
\langle \Sq^r_* \chi \xi_i, x \rangle &= \langle \chi \xi_i, \Sq^r x \rangle \\
&= \langle \chi \xi_i, \mu(\Sq^r \otimes x) \rangle \\
&= \langle \xi_i, \chi \mu(\Sq^r \otimes x) \rangle \\
&= \langle \xi_i, \mu(\chi x \otimes \chi \Sq^r) \rangle \\
&= \langle \Delta \xi_i, \chi x \otimes \chi \Sq^r \rangle \\
&= \langle \sum_k \xi_{i-k}^{2^k} \otimes \xi_k, \chi x \otimes \chi \Sq^r \rangle \\
&= \sum_k \langle \xi_{i-k}^{2^k}, \chi x \rangle \cdot \langle \xi_k, \chi \Sq^r \rangle \\
&= \sum_k \langle \chi \xi_{i-k}^{2^k}, x \rangle \cdot \langle \chi \xi_k, \Sq^r \rangle \\
&= \sum_k \langle \chi \xi_{i-k}^{2^k}, x \rangle \cdot \delta_{r=2^k-1} \\
&= \begin{cases} \langle \chi \xi_{i-k}^{2^k}, x \rangle, & \text{if } r = 2^k - 1 \\ 0, & \text{otherwise}. \end{cases}
\end{align}