Homotopy group of a spectrum $\{ E_n,s_n\}$ as the colimit of $\pi_{k+N}(E_{N})$

I want to prove that the following two definitions of homotopy group of a spectra are equivalent:

DEF 1 Let $S= \Sigma^{\infty}S^0$ be the spectrum of spheres. The group $[\Sigma^kS,E]$ is called the k-th homotopy group of $E$ and denoted by $\pi_k(E)$.

and the second

DEF 2 $\pi_k(E)=\varinjlim_{N \to \infty} \pi_{k+N}(E_N)$ where the directed system is build via the following composition: $$\pi_{k+N}(E_N \ast)\xrightarrow{\Sigma} \pi_{k+N+1}(\Sigma E_N, \ast) \xrightarrow{(s_N)_{\ast}} \pi_{k+N+1}(E_{N+1}, \ast)$$

My Attempt (based on Switzer)

observe that the cofinal subspectra of $\Sigma^k S$ are the spectra $\Sigma^{k-r}\Sigma^{\infty}S^r$, for $r\geq 0$. A map $f\colon \Sigma^{k-r}\Sigma^{\infty} S^r \to E$ is just a function $f\colon S^r \to E_{r-k}$ together with all its suspensions $\Sigma^sf \colon S^{r+s}\to \Sigma^s E_{r-k}\subset E_{r+s-k}$. The groups $\pi_{k+n}(E_k,\ast)$ where $k\geq \min (2,2-n)$ and morphisms $$\pi_{k+n}(E_n \ast)\xrightarrow{\Sigma} \pi_{k+n+1}(\Sigma E_n, \ast) \xrightarrow{(s_n)_{\ast}} \pi_{k+n+1}(E_{n+1}, \ast)$$ define a direct system of abelian groups, and we can define \begin{align*} \alpha \colon [\Sigma^nS^0,E] &\to \varprojlim_{N \to \infty} \pi_{k+N}(E_N) \\ \{ \Sigma^{k-r}\Sigma^{\infty}S^r , f \} &\mapsto \{[f]\} \end{align*} where $\{ \Sigma^{k-r}\Sigma^{\infty}S^r , f \}$ means that $f$ is defined only on $\Sigma^{k-r}\Sigma^{\infty}S^r$ since it is a morphism of spectra. Switzer claims that $\alpha$ is surjective, but I can't show that. Being directed the index category, it's enough to build a preimage for $[g \colon S^{m+k} \to E_{m}]$ where the equivalence relation is the one defining the colimit of abelian groups. (See here)

But if I take the map of spectra $\Sigma^{-m}\Sigma^{\infty}S^{m+k} \to E$ defined by $g$, I can't control the image of the maps in higher degree.

It is supposed to be trivial, but I can't see how

I think you are interpreting wrongly the map suggested in Switzer. My idea is that you have a map of spectra $\bar{f} \colon \Sigma^{k-r}\Sigma^{\infty}S^r \to E$ and then $\alpha$ assigns to it the map $f\colon S^r\to E_{r-k}$. Note that bar over $f$ before, it's a map of spectra (and a morphism from $\Sigma^kS$)!
So surjectivity is indeed trivial, in fact for a given $[g \colon S^{m+k} \to E_{m}]$, you just define the map $$\bar{g}\colon \Sigma^{-m}\Sigma^{\infty}S^{m+k} \to E$$ in the canonical way (you take $g$ and its suspensions) and $\alpha(\bar{g})=[g]$ by definition.