Suppose I have a CW complex $Y$ of dimension $n+2$ and let $X_{n+2}$ be the third nontrivial Postnikov stage of $S^n$ (i.e. there is a map $S^n \to X_{n+2}$ which is an $(n+2)$equivalence). We obtain a fibration $F\to X_{n+2} \to K(\mathbb Z, n)$ (which factors through $X_{n+1}$). Thus we have $$ K(\mathbb Z,n1) \to F \to X_{n+2} \to K(\mathbb Z,n) $$ and therefore the exact sequence $$ H^{n1}(Y;\mathbb Z) \to [Y,F] \to [Y,X_{n+2}] \to H^n(Y;\mathbb Z). $$ I would like to understand the map $H^{n1}(Y;\mathbb Z) \to [Y,F]$. Is there a good model for $F$ such that $[Y,F]$ is expressed by some topological data of $Y$? Clearly $F$ is the homotopy fiber of $\text{Sq}^2 \colon K(\mathbb Z_2,n+1) \to K(\mathbb Z_2,n+3)$.

10$\begingroup$ The first three layers of the Postnikov tower of $S^n$ are determined by the homotopy groups $\mathbb Z,\mathbb Z/2,\mathbb Z/2$, the $k$invariants $k_n:K(\mathbb Z,n)\to K(\mathbb Z/2,n+2)$ and $k_{n+1}:K(\mathbb Z/2,n+1)\to K(\mathbb Z/2,n+3)$, and a nullhomotopy of $\Sigma k_{n+1}\circ k_n$. As you say, $k_{n+1} = Sq^2$, whereas $k_n$ is the composition of mod 2 reduction with $Sq^2$. The nullhomotopy comes from the Adem relation $Sq^2Sq^2 = Sq^3Sq^1$ and the fact that $Sq^1$ vanishes on mod 2 reductions (Bockstein exact sequence). One can probably work this out explicitly on cochains... $\endgroup$ – Bertram Arnold Jan 31 '19 at 16:01
Let me treat the stable case ($n\ge 4$), so that we are interested in the first three stages of the Postnikov tower of the sphere spectrum. The way the Postnikov tower works is that if the homotopy groups of some spectrum $E$ are concentrated in degrees $0$ to $n+1$, we get a cofiber sequence $\Sigma^{n+1} H\pi_{n+1}(E)\to E\to \tau_{\le n}E$, where $\tau_{\le n}E$ has homotopy groups concentrated in degrees $0$ to $n$, and the map from $E$ is an isomorphism in these degrees. By rotating this cofiber sequence, we obtain the nth kinvariant $k_{n}:\tau_{\le n}E\to \Sigma^{n+2} H\pi_{n+1}(E)$, and $E$ can be recovered as the fiber of this map. Thus the extension of the Postnikov tower to the next stage is determined by $[\tau_{\le n}E,\Sigma^{n+2}H\pi_{n+1}(E)] = H^{n+2}(E;\pi_{n+1}(E))$.
For $n=0$, this is completely trivial: All spectra with homotopy groups concentrated in one degree are EilenbergMacLane spectra.
For $n=1$, we have two groups $\pi_0(E),\pi_1(E)$ and $k_0\in H^2(H\pi_0(E);\pi_1(E))$. For any group $A$, we have $H_1(HA;\mathbb Z) = 0,H_2(HA;\mathbb Z)\cong A\otimes\mathbb Z/2$, so by the UCT, we have an isomorphism $H^2(H\pi_0(E);\pi_1(E))\cong \operatorname{Hom}(\pi_0(E)\otimes \mathbb Z/2,\pi_1(E))$. Now precomposition with $\eta\in \pi_1(S)\cong \mathbb Z/2$ also determines a map $\pi_0(E)\otimes\mathbb Z/2\to \pi_1(E)$, and this is precisely the homomorphism corresponding to the $k$invariant: By the Yoneda lemma we only have to check this for the (second stage of the Postnikov tower of) the sphere spectrum which has nontrivial 0th $k$invariant (otherwise $H_1(S)\cong\mathbb Z/2$) and multiplication by $\eta$, so both homomorphisms are the nontrivial map from $\mathbb Z/2$ to itself.
In particular, from $\eta^2\neq 0$ we obtain that the kinvariants of $\tau_{\le 1} S$ and $\tau_{\ge 0,\le 1}\Sigma^{1}S$ are nontrivial. We now have to determine how these two layers of the Postnikov tower fit together. Recall that $k_1\in H^3(\tau_{\le 1}S;\mathbb Z/2)$; from the (co)fiber sequence $\Sigma^1 H\mathbb Z/2\to \tau_{\le 1}S\to\mathbb Z$ we obtain a long exact sequence $$ H^1(H\mathbb Z/2;\mathbb Z/2)\to H^3(H\mathbb Z;\mathbb Z/2)\to H^3(\tau_{\le 1}S;\mathbb Z/2)\to H^2(H\mathbb Z/2;\mathbb Z/2)\to H^4(H\mathbb Z;\mathbb Z/2) $$ The mod 2 reduction map $H^*(H\mathbb Z/2;\mathbb Z/2)\to H^*(H\mathbb Z;\mathbb Z/2)$ is the quotient by the image of precomposition with $Sq^1$. The maps in this long exact sequence can be factored as $H^*(H\mathbb Z/2;\mathbb Z/2)\xrightarrow{\cdot Sq^2} H^{*+2}(H\mathbb Z/2;\mathbb Z/2)\to H^{*+2}(H\mathbb Z;\mathbb Z/2)$ since 0th kinvariant of $S$ is nontrivial. Thus this long exact sequence becomes $$ \mathbb Z/2\langle Sq^1\rangle\to \mathbb Z/2\langle Sq^3\rangle\to H^3(\tau_{\le 1}S;\mathbb Z/2)\to \mathbb Z/2\langle Sq^2\rangle \to \mathbb Z/2\langle Sq^4\rangle $$ So the first map sends $Sq^1$ to $Sq^1Sq^2 = Sq^3$, so it is an isomorphism. The last map is $Sq^2\mapsto Sq^2Sq^2 = Sq^3Sq^1$ which gets sent to $0$ in the quotient. Thus we obtain that the restriction $H^3(\tau_{\le 1}S;\mathbb Z/2)\to H^3(\Sigma^1H\mathbb Z/2;\mathbb Z/2)$ is an isomorphism. This means that the first $k$invariant of $\tau_{\le 2}S$ is uniquely determined by its restriction, the $0$th kinvariant of $\tau_{\ge 0,\le 1}\Sigma^{1}S$, which we have seen to be nontrivial.
As to your question about the natural transformation $H^{n1}(Y)\to [Y,F]$, by the Yoneda lemma we only have to write down a map $K(\mathbb Z,n1)\to F$, which by the definition of $F$ as a homotopy fiber is essentially the same as a cochain with boundary $Sq^2Sq^2 \iota_{n1}\in H^{n+3}(K(\mathbb Z,n1);\mathbb Z/2)$. Such a cochain does indeed exist, essentially by the above discussion, but I think it would be pretty hard to write down explicitly (since the same is true for the Steenrod operations), and it's not really clear to me what the value of such a construction would be. Maybe it would help if you can explain what a sufficiently good model would be for you?