$d^3$ in the Atiyah-Hirzebruch spectral sequence for (twisted) $KO$

Here are the first, more straightforward, differentials in the AHSS $$H^p(X;KO^q(\ast)) \Rightarrow KO^{p+q}(X)$$ for real K-theory. Note $KO^q(\ast)$ is $$ \begin{cases} \Bbb Z &\text{if }q = 8k,\\ \Bbb Z/2 &\text{if }q = 8k-1,\\ \Bbb Z/2 &\text{if }q = 8k-2,\\ \Bbb Z &\text{if }q = 8k-4,\\ 0 &\text{otherwise.} \end{cases} $$

• For $q=8k$ the map $$d_2: H^\ast(X;\Bbb Z) \to H^{\ast+2}(X;\Bbb Z/2)$$ is $Sq^2 \circ r_2$, where $r_2$ is reduction from integer cohomology to mod-2 cohomology.

• For $q=8k-1$ the map $$d_2: H^\ast(X;\Bbb Z/2) \to H^{\ast+2}(X;\Bbb Z/2)$$ is $Sq^2$. (These two differentials are what Denis Nardin alluded to in the comments.)

• For $q=8k-2$ the map $$d_3: H^\ast(X;\Bbb Z/2) \to H^{\ast+3}(X;\Bbb Z)$$ is $\beta_2 \circ Sq^2$, where $\beta_2$ is the Bockstein from mod-2 cohomology to integer cohomology.

• For $q=8k-4$ the map $$d_5: H^\ast(X;\Bbb Z) \to H^{\ast+5}(X;\Bbb Z)$$ is $\beta_2 \circ Sq^4 \circ r_2 + \beta_3 \circ P^1 \circ r_3$, where $P^1$ is a Steenrod operation on mod-3 cohomology. (Or perhaps this is $-d_5$, depending on choice of generators.)

The only remaining $d_3$ differential, from $E_3^{p,8k}$ to $E_3^{p+3,8k-2}$, is given by a map $$ \Phi_{1,1} \circ r_2: \ker(Sq^2 \circ r_2) \to \mathrm{coker}(Sq^2). $$ This operation $\Phi_{1,1}$ is a so-called secondary cohomology operation, associated to the relation $Sq^2 Sq^2 + Sq^3 Sq^1 = 0$ between cohomology operations. There is no straightforward description of $\Phi_{1,1}$ in terms of primary operations. In the same way that $Sq^1$ in $H^*(X)$ detects the presence of degree-2 attaching maps $S^n \to S^n$ among the cells of $X$, and $Sq^2$ detects the presence of the Hopf map $\eta$ in an attaching map $S^{n+1} \to S^n$, the operation $\Phi_{1,1}$ detects the presence of $\eta \circ \eta$ in an attaching map $S^{n+2} \to S^n$.

I am not certain of the twisted K-theory versions of some of these, but this is a deficiency in my knowledge of real twisted K-theory.

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.