Expressions for $d^2$ and $d^3$ of the twisted Atiyah-Hirzebruch spectral sequence can be found in Grady, Daniel, and Hisham Sati. "Twisted differential KO-theory." arXiv preprint arXiv:1905.09085 (2019) Proposition 18, where the twist is given by elements $(\sigma_1,\sigma_2)\in H^1(X,\mathbb Z/2)\times H^2(X,\mathbb Z/2)\cong H^*(X,\tau_{\le 1}bgl_1 KO)$:

- The $E_2$-page is given by twisted cohomology $H^p(X;(KO^q)_{\sigma_1})$, with $\mathbb Z/2$ acting by the sign representation on all groups
- The only nonvanishing $d_2$ are
\begin{align*}
d_2^{p,-8t}:H^p(X;\mathbb Z_{\sigma_1})\to H^{p+2}(X;\mathbb Z/2)&, x\mapsto \operatorname Sq^2(r(x)) + \sigma_1\cup Sq^1(r(x)) + \sigma_2\cup r(x)\\
d_2^{p,-8t-1}:H^p(X;\mathbb Z/2)\to H^{p+2}(X;\mathbb Z/2)&, x\mapsto \operatorname Sq^2(x) + \sigma_1\cup Sq^1(x) + \sigma_2\cup x
\end{align*}
- The only canonically defined $d_3$ is
$$
d_3^{p,-8t-2}:H^p(X;\mathbb Z/2)\to H^{p+2}(X;\mathbb Z_{\sigma_1}), x\mapsto \operatorname \beta(Sq^2(x)) + \beta(\sigma_2\cup r(x))\\
$$

Here $r: H^p(X;\mathbb Z_{\sigma_1}) \to H^p(X;\mathbb Z/2)$ and $\beta: H^p(X;\mathbb Z/2) \to H^{p+1}(X;\mathbb Z_{\sigma_1})$ are reduction mod 2 and the corresponding Bockstein.

The above twists are those defined by geometry, i.e. by the inclusion $\operatorname {sLine}^\otimes\to \operatorname {Pic}(\operatorname{sVect}^\otimes)$ of superlines as tensor-invertible super vector spaces, i.e. the twists arising from ``super gerbes'' on $X$. More generally, twists for $KO$-theory are classified by $H^*(X;bgl_1 KO)\cong H^1(X,\mathbb Z/2)\times H^2(X,\mathbb Z/2)\oplus H^*(X,\tau_{> 1}bgl_1 KO)$. By degree considerations, you can see that the ones in the third group don't change the above differentials (for $d_3$, use that the multiplication $KO_2\otimes KO_{8t+2}\to KO_{8t+4}$ vanishes).

The only remaining canonically defined differential is
$$
d_5^{p,-8t-4}:H^p(X;\mathbb Z_{\sigma_1})\to H^{p+4}(X;\mathbb Z_{\sigma_1})
$$
I don't know anything about it besides the untwisted part, which you can find in Tyler Lawson's answer.