Computation of stable homotopy groups of $RP^2$ 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 $\pi_2^s(RP^2)$?
 A: Another approach will be to use Adams Spectral Sequence. This is a most suited to compute stable homotopy groups. An outline will be as follows. 
As a module over steenrod algebra $\mathbb{RP}^{2}$ looks like two cells connected with $Sq^{1}$. 
One can either compute the minimal resolution or do a May spectral sequence to compute the $E_{2}$ page of Adams Spectral Sequence. In fact, May Spectral Sequence is tailor made for such modules. All one has to do is run May SS after throwing away $h_{0}$ But after first few terms you want to probably want to see if you have any differentials in the Adams SS.
If you are going high enough to worry about Adams SS differentials, one can import the computations done for Sphere spectrum. Since $\mathbb{RP}^{2}$ is cofiber of $S \xrightarrow{2} S$ and multiplication by $2$ induces multiplication by $h_{0}$ in Adams SS. Also one has a long exact sequence 
 $Ext_{A}^{s,t}(S) \xrightarrow{h_{0}} Ext_{A}^{s+1,t+1}(S) \to Ext_{A}^{s+1, t+1}(\mathbb{RP}^{2}) \to Ext_{A}^{s+1, t}(S)$ 
where $A$ is the mod 2 Steenrod Algebra. Using this method one can import the differentials Adams SS differentials for $S$, the sphere spectrum. One can easily recover homotopy groups $\mathbb{RP^{2}}$ from the knowledge of sphere. This obviously gives the answer easily up to $50^{th}$ stem, may be more, but I did not go beyond that. `
This approach will give all the extensions as well. Also for $\mathbb{RP}^{2}$ one has to worry about what happens at prime $2$ only.  `
A: 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.
