moving from sphere spectrum to finite spectrum

Let $\{Y,Z\}$ denote the homotopy group of maps between a spectrum $Y$ and a spectrum $Z$. We are assuming that one has a sequence of spectra $Z^1, Z^2, \dots$ such that for any fixed $i$, $\{S^i,Z^k\} = 0$ for $k>>0$, and we want to show that, if $Y$ is a finite spectrum, then $\{Y,Z^k\} = 0$ for $k>>0$.

An easy way to see this is by induction on the number of cells of $Y$, with inductive step as follows: if $Y$ is obtained from $X$ by attaching an $n$--cell, then, by basic homotopy theory, the cofibration sequence $X \rightarrow Y \rightarrow S^n$ induces $\{S^n,Z^k\} \rightarrow \{Y,Z^k\} \rightarrow \{X,Z^k\}$ exact in the middle. By assumption and inductive hypothesis, the first and last of these are 0 for $k>>0$, and thus the so is the middle term. [This is clearer than Hatcher, who is sort of saying the same thing much more awkwardly.]

For convenience I will set $Z = Z^k$, where $Z^k$ is as in your notation.

Case 1: $Y$ is the suspension spectrum of a based finite complex $U$ having dimension $n$. Then $\pi^Y_\ast(Z)$ is given by the homotopy groups of the function space of based maps $F(U,\Omega^\infty Z)$, where $\Omega^\infty Z$ is the infinite loop space associated with $Z$.

Filter $U$ by its skeleta $U^{(j)} \subset U$. Then we have a sequence of fibrations $$ F(U,\Omega^\infty Z) = F(U^{(n)},\Omega^\infty Z) \to F(U^{(n-1)},\Omega^\infty Z)\to \cdots \to F(U^{(0)},\Omega^\infty Z)\, . $$ The fibers of the map at stage $j$ in the sequence are identified with $F(U^{(j)}/U^{(j-1)},\Omega^\infty Z)$ and the latter is a finite product of copies of $F(S^j,\Omega^\infty Z)$.

The homotopy groups of $F(S^j,\Omega^\infty Z)$ are just the homotopy groups of $Z$ shifted by $j$. So, by your assumptions, for any index $i$ there is a $k$ with $Z= Z^k$ such that $\pi_\ell$ of $F(S^j,\Omega^\infty Z)$ vanishes for $\ell\le i$. Consequently, the fiber of the $j$th fibration is $i$-connected. Furthermore, a similar argument shows that $F(U^{(0)},\Omega^\infty Z)$ is $i$-connected. It follows that $F(U,\Omega^\infty Z)$ is also $i$-connected.

Case 2: $Y$ is an arbitrary finite spectrum. Then there is a non-negative integer $t$ such that $\Sigma^t Y$ is the suspension spectrum of a finite complex $U$. So, $F(Y,Z) = F(U,\Sigma^t Z)$. The argument then works as above with $Z$ replaced by $\Sigma^t Z$.

Remark: What the above in effect shows is this: let $Y$ be a finite dimensional cell spectrum of dimension $s\in \Bbb Z$ and let $Z$ be an $r$-conected spectrum. Then the function spectrum $F(Y,Z)$ is $(r-s)$-connected.