This is a partial answer designed to help you understand the relation between killing $\pi_3$ and the appearance of a $K(\mathbb{Z},2)$. It was way too long for a comment, but I thought it might be useful to you.
For $n=1$, $Spin(1)\cong\mathbb{Z}_2$ and for $n=2$ $Spin(2)\cong S^1$. In these cases the string groups as you desribe them don't really exist, so we'll assume $n\geq 3$ from now on.
Then $Spin(3)\cong S^3\cong SU(2)$ so $String(3)\simeq S^3\langle 3\rangle$ is just the $3$-connected cover of $S^3$. All the homotopy groups of this space are torsion, by a Theorem of Serre, and it has non-trivial homotopy groups in infinitely many degrees.
On the other hand $Spin(4)\cong S^3\times S^3$ is not simple, so again $String(4)$ is a special case. To avoid these difficulties we'll assume that $n>4$ in the following. Note that what we say will still apply to $Spin(3)$, but if care is taken with statements (importantly, for example, the inclusion $Spin(3)\hookrightarrow Spin(n)$ does not induce an isomorphism on $\pi_3$).
Now for $n>4$, $Spin(n)$ is a simply connected, simple compact Lie group. Therefore it holds that $\pi_1Spin(n)=0=\pi_2Spin(n)$ and $\pi_3Spin(n)\cong\mathbb{Z}$. The Hurewicz theorem tells us that $H_2Spin(n)=0$ and $H_3Spin(n)\cong \pi_3Spin(n)\cong\mathbb{Z}$, and the universal coefficient theorem then tells us that $H^3Spin(n)\cong Hom(H_3Spin(n),\mathbb{Z})\cong \mathbb{Z}$. Let $x_3\in H^3Spin(n)$ be a generator. Then this cohomology class is represented by a map
$$x_3:Spin(n)\rightarrow K(\mathbb{Z},3)$$
into the integral Eilenberg-Mac Lane space in degree $3$. This cohomology class is primitive for degree reasons so its homotopy fibre is an H-space. It will actually turn out that $x_3$ is the cohomology suspension of a universal class $c^s_2\in H^4BSpin(4)$. Therefore $x_3$ is a loop map and its homotopy fibre has the homotopy type of a topological group, which is homotopy equivalent to $String(n)$. In any case we have a homotopy fibration sequence
$$String(n)\rightarrow Spin(n)\xrightarrow{x_3} K(\mathbb{Z},3)$$
which represents a delooping of the sequence written in your question. This is how $K(\mathbb{Z},2)\simeq \Omega K(\mathbb{Z},3)$ appears in your sequence.
To understand what is going on it may help to first note that the complexification map $SU(2)\rightarrow SO(n)$ may be lifted to the universal cover as a homomorphism $i:SU(2)\cong S^3\rightarrow Spin(n)$ whose homotopy class generates $\pi_3Spin(n)$, and this homomorphism classifies to a map $Bi:BSU(2)\rightarrow BSpin(n)$. We know $\pi_4BSpin(n)\cong H^4BSpin(n)\cong \mathbb{Z}$ are the first nontrivial homotopy/cohomolgy groups, so for an appropriate choice of generators $c^s_2\in H^4BSpin(4)$ it holds that $Bi^*c_2^s=c_2$, where $c_2\in H^4BSU(2)$ is the 2nd Chern class. This follows since $Bi$ induces an isomorphism on $\pi_4$.
The point is that $x_3\simeq \Omega c_2^s$ so there is another homotopy fibration sequence
$$BString(n)\rightarrow BSpin(n)\xrightarrow{c_2^s} K(\mathbb{Z},4)$$
which returns the previous sequence I wrote above upon applying $\Omega$.
The main point to understand is that this a general procedure. For any simply connected space $X$ and any spherical homology class $a\in H_n(X;A)$ in the image of the Hurewicz map $\pi_nX\rightarrow H_nX$. Choose $\alpha\in \pi_n$ representing $a$ in this sense. Then $a$ generates a subgroup $A\leq H_nX$ and so corresponds to a class in $H^n(X;A)$. Then there is a homotopy fibre sequence
$$X\langle\alpha\rangle\rightarrow X\xrightarrow{x}K(A,n)$$
defining the space $X\langle\alpha\rangle$ which is obtained by killing the class $\alpha$. This is obviously clearest when $X$ is, say, $n$-connected.
