Your first question can be rephrased as follows.
Can we find an $(n-1)$-connected space $Y$ such that there is a bijective correspondence between $H^n(K^{n+1};\Bbb Z)$ and $[K^{n+1}, Y]$ for any $(n+1)$-dimensional CW complex $K^{n+1}$?
The short answer to this question is yes. You can take the Eilenberg-MacLane space $Y=K(\Bbb Z,n)$. This gives you a natural isomorphism $[X,K(\Bbb Z,n)]\cong H^n(X;\Bbb Z)$ when $X$ is any CW complex, not necessarily $(n+1)$-dimensional.
However, $K(\Bbb Z,n)$ is not easy to describe in general, so I guess it makes sense to ask the following question.
If we only care about $(n+1)$-dimensional complexes, do we have a choice for $Y$ that is easier to describe?
Yes. We can take $Y=S^1$ if $n=1$ and $Y=\Sigma^{n-2}\Bbb CP^2$ [the $(n-2)$-fold reduced suspension of $\Bbb CP^2$] if $n\ge 2$.
Assume $n\ge 2$ since $S^1$ is a $K(\Bbb Z,1)$. Let $Y$ be built from $S^n$ by attaching $(n+2)$-cells which kill elements in $\pi_{n+1}(S^n)$. More explicitly, for $n\ge 2$, $\pi_{n+1}(S^n)$ is generated by the $(n-2)$-fold suspension of Hopf map $\eta$, so $Y=C(\Sigma^{n-2}\eta)$ is the mapping cone of $\Sigma^{n-2}\eta$. By considering the long exact sequence of homotopy groups associated with the pair $(Y,S^n)$, we see that $Y$ is $(n-1)$-connected, $\pi_n(Y)\cong\pi_n(S^n)$, and $\pi_{n+1}(Y)=0$.
Now, we can attach cells of dimension at least $n+3$ to kill higher homotopy groups of $Y$ and "eventually" arrive at $K(\Bbb Z,n)$. So $Y$ is essentially the $(n+2)$-skeleton of $K(\Bbb Z,n)$.
We now claim that $$H^n(K^{n+1};\Bbb Z)\cong[K^{n+1},K(\Bbb Z,n)]\overset{\sim}{\underset{i_\ast}{\longleftarrow}}[K^{n+1},K(\Bbb Z,n)^{(n+2)}]=[K^{n+1},Y]$$
It suffices to prove the bijection in the middle. The map $i_\ast:[K^{n+1},Y]\to [K^{n+1},K(\Bbb Z,n)]$ is induced by the inclusion $i:Y\to K(\Bbb Z,n)$. The surjectivity of $i_\ast$ follows directly from of cellular approximation theorem; injectivity follows from the fact that if $H:K^{n+1}\times I\to K(\Bbb Z,n)$ is a homotopy between two (cellular) maps then cellular approximation (for pairs) implies that $H$ is homotopic to a cellular map $G:K^{n+1}\times I\to K(\Bbb Z,n)^{(n+2)}=Y$ rel $K^{n+1}\times \{0,1\}$.
Therefore, we get a natural bijection between $H^n(K^{n+1};\Bbb Z)$ and $[K^{n+1}, Y]$, where $Y=C(\Sigma^{n-2}\eta)$. We have $C(\Sigma^{n-2}\eta)\cong \Sigma^{n-2}C(\eta)\cong\Sigma^{n-2}\Bbb CP^2$, so the natural bijection can be written as
$$[K^{n+1},\Sigma^{n-2}\Bbb CP^2]\cong H^n(K^{n+1};\Bbb Z)$$
This works when $K^{n+1}$ is any CW complex with dimension at most $n+1$ and $n\ge 2$.
The answer to your 2nd question is that we always have a surjection from $[K^{n+1},S^n]$ to $H^n(K^{n+1};\Bbb Z)$ for any $(n+1)$-dim complex. Note that the $(n+1)$-skeleton of $K(\Bbb Z,n)$ is $K(\Bbb Z,n)^{(n+1)}=S^n$ because no $(n+1)$-cells are attached in the process of killing higher homotopy groups. By cellular approximation theorem, any continuous map $f:K^{n+1}\to K(\Bbb Z,n)$ is homotopic to a cellular map $\tilde{f}:K^{n+1}\to K(\Bbb Z,n)^{(n+1)}=S^n$, so the map $$j_\ast:[K^{n+1},S^n]\longrightarrow [K^{n+1},K(\Bbb Z,n)]\cong H^n(K^{n+1};\Bbb Z)$$ induced by the inclusion $S^n\subseteq K(\Bbb Z,n)$ is surjective.
