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I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

Let $A=H_1(X,\mathbb{Z}/2)$. We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. This actually holds for any $Y=K(A,1)$ with $A$ abelian. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because $A=H_1(X,\mathbb{Z}/2)$, so it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

Let $A=H_1(X,\mathbb{Z}/2)$. We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. This actually holds for any $Y=K(A,1)$ with $A$ abelian. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because $A=H_1(X,\mathbb{Z}/2)$, so it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

We can also proceed without using the Brown-Loday paper, as hinted below by Tom Goodwillie in a comment. It suffices to show that $\pi_2^{st}(Y)$ is a $\mathbb{Z}/2$-vector space. We have $\pi_2^{st}(Y)=\pi_4(\Sigma^2Y)$ since $Y$ is connected. The space $Y$ is a product of copies of $\mathbb{R}P^\infty$. By the splitting of the suspension of a product, $\Sigma^2 Y$ is a wedge of copies of $\Sigma^2(\mathbb{R}P^\infty\wedge\stackrel{n}\cdots\wedge\mathbb{R}P^\infty)$. The latter space is $4$-connected for $n>2$, hence $\pi_2^{st}(Y)$ is a direct sum of copies of $\pi_4\Sigma^2\mathbb{R}P^\infty=\mathbb{Z}/2$ and $\pi_4\Sigma^2(\mathbb{R}P^\infty\wedge\mathbb{R}P^\infty)=\mathbb{Z}/2$.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

Let $A=H_1(X,\mathbb{Z}/2)$. We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. This actually holds for any $Y=K(A,1)$ with $A$ abelian. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because $A=H_1(X,\mathbb{Z}/2)$, so it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.

I love this question! I've enjoyed thinking of it.

By naturality, the sequence splits always if and only if it splits for $X=K(A,1)$ with $A$ an abelian group ($A=H_1(X,\mathbb{Z})$). Actually, the sequence splits for $X=K(A,1)$ if and only if it splits for all spaces with $H_1(X,\mathbb{Z})=A$. In case $X=K(A,1)$, we have an isomorphism of short exact sequences: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_2(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

Now the problem is reduced to a purely algebraic one. It is easy to see that the top sequence splits in many cases, e.g.~if $A$ is finitely generated, using the structure theorem. Actually, $A\hat{\otimes}A$ and $A\wedge A$ are quadratic functors on $A$, both with cross effect $\otimes$, the tensor product. Moreover, the quotient natural map $A\hat{\otimes}A\to A\wedge A$ induces an isomorphism between cross effects. Therefore, the class of abelian groups for which the sequence splits is closed under direct sum. This class includes cyclic groups, $2$-divisible groups, and the Pruffer group $\mathbb{Z}/2^\infty$ (the 2-primary component of $\mathbb{Q}/\mathbb{Z})$. In all the previous cases, one of the terms of the short exact sequence vanishes. Hence, the class also includes, finitely generated groups, torsion groups, etc.

Unfortunately my knowledge of infinite abelian groups is not sufficient to find a counterexample (or a complete proof).

PS. What happens for $A=\mathbb{Z}_2$ the $2$-adic integers?

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.