I'm following the book of Davis and Kirk "Lecture notes in Algebraic Topology" where, at pages $246-247$ there is a computation of the aforementioned groups via the Atiyah- Hirzebruch Spectral Sequence.
Using the well-known fact that $\Omega^{SO}_k(*)\cong \mathbb{Z}$ for $k= 0,4$, and it is trivial for $k=1,2,3$ together with the split injevtiveness of the edge homomorphism it's easy to conclude that for $n\leq 3$, $\Omega^{SO}_n(X)\cong H_n(X)$.
I'm interested in computing the $4$th Group, since in this degree there is a possible non-trivial extension problem. At the stable page, in position $(0,4)$ there is $\mathbb{Z}$, in position $(4,0)$ there is $H_4(X)$. If the position of these two groups were exchanged, by the projectivity of $\mathbb{Z}$ (as an abelian Group) the extension problem would be trivial. But in this case, since it is not injective I dont know how to deal with the Short exact sequence.
According to the reference it should be a trivial extension problem, i.e. $\Omega^{SO}(X)_4\cong H_4(X)\oplus \mathbb{Z}$ but they dont give any reason for it.
Can someone provide some explanations for this?
You're right that we have an exact sequence $0 \to \Bbb Z \to \Omega^{SO}_4(X) \to H_4(X) \to 0$. Here are a couple of ways we can show this sequence splits.
1) Identify the first map as $\Omega^{SO}_4(pt) \to \Omega_4^{SO}(X)$ for some choice of basepoint. Then there is an obvious splitting given by retracting to that basepoint. I suspect this is the solution they intended, since they note above your result that $\Omega_n(pt) \to \Omega_n(X)$ is a split injection for any $X$.
2) The right map is the map that forgets the manifold defining it. You can get an inverse to it by picking a choice of manifold representing it of signature 0. (This is sorta cheating, since I would only think of it if I already knew what the first map was.)