I would like to compute the first few stable homotopy groups of $RP^2$.

I first thought to use the Atiyah-Hirzebruch Spectral Sequence, (see Davis & Kirk, pg. 242). Here is what I computed for the $E^2$ term of the spectral sequence: $$E^2_{p,q}=\begin{array}{|ccc} \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\ \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\ \mathbb{Z} & \mathbb{Z}_2 & 0 \\\hline \end{array}$$

From this, I compute that the associated graded complex to $\pi_1^s(RP^2)$ is $\mathbb{Z}_2\oplus\mathbb{Z}_2$. (I think I made a mistake here with the local coefficients. I believe I showed the local coefficients act trivially, so it should just reduce to ordinary homology with coefficients in $\pi_q^s(S^0)$.) So either $\pi_1^s(RP^2)$ is $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2$.

On the other hand, we know that $\pi_1^s(RP^2)=\pi_2(\Sigma RP^2)$ by the Freudenthal suspension theorem. Using the evident cell structure on $\Sigma RP^2$ consisting of a single 0-cell, a single 2-cell, and a single 3-cell, we see that $\pi_1(\Sigma RP^2)=0$ by cellular approximation. So by the Hurewicz theorem $\pi_2(\Sigma RP^2)\cong H_2(\Sigma RP^2) \cong H_1(RP^2) \cong \mathbb{Z_2}$.

Where am I going wrong using the AHSS? How does one compute $\pi_2^s(RP^2)$?