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 $\pi\_2^s(RP^2)$?



Computation of stable homotopy groups of $RP^2$
The version of the AHSS you wrote down converges to $\pi\_\*^s(\mathbb{RP}^2\_+)$, i.e. with an extra basepoint. This splits canonically as $\pi\_\*^s(\mathbb{RP}^2) \oplus \pi\_\*^s(S^0)$, and the left hand column of your chart corresponds to the second summand. Thus, to get $\pi\_\*^s(\mathbb{RP}^2)$ you should throw that column away.


