Splitting of Singular Homologies In Singular homology, let $C_n(X)$ be the free abelian group generated by all the $n$-siimplices of the topological space $X$. Let $U$ be a subspace of $X$, then we have a spliting sequence 
$0\rightarrow C_n(U) \rightarrow C_n(X) \rightarrow C_n(X,U) \rightarrow 0$.
But why we do not have the splitting $H_n(X) \cong H_n(U) \oplus H_n(X,U)$ of homologies? 
 A: You're right that for any pair $(X, A)$, there is a split short exact sequence of chain complexes
$$0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$
But it is not true that having short exact sequences at the chain level implies that the snake maps $\partial$ at the homology levels are zero, i.e., you get a short exact sequence at the homology level. This is because a chain-level injective/surjective map might not induce an injective/surjective map at the homology level, see this question. 
The point of the long exact sequence of homology is that it measures failure of short exactness of the $H_\bullet$ functors, to emphasize.

One a different note, if there is a retract $r : X \to A$, then the induced $H_\bullet(r) : H_\bullet(X) \to H_\bullet(A)$ acts as a left-inverse for the maps $H_\bullet(X) \to H_\bullet(A)$, making the snake maps vanish and becomes a section of the resulting short exact sequence 
$$0 \to H_\bullet(A) \to H_\bullet(X) \to H_\bullet(X, A) \to 0$$
Which implies $H_\bullet(X) \cong H_\bullet(A) \oplus H_\bullet(X, A)$.
A: Adding an answer here that I found helpful when thinking about why the homology groups split in the case of retractions: a retraction $r:X \to A$ is left inverse for the inclusion $i$. Functors always preserve mutual left/right inverse-ness (in other words they preserve split epi/mono pairs) so there is a left inverse for the induced map $H_\bullet(i):H_\bullet(A) \to H_\bullet(X)$. On the other hand, the pointwise splitting of a chain complex is not necessarily a chain-map. More precisely, the short exact sequence of chain groups
$$ 0 \to C_n(A) \to C_n(X) \to C_n(X, A) \to 0$$ splits in $\mathrm{Ab}$ for any given $n$, but the short exact sequence of chain-complexes
$$ 0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$ may not split in the category of chain complexes. If the latter sequence splits (such as when there is a retraction $r:X \to A$, or when $X$ is the direct sum of $A$ and some other space), then by applying the homology functor you do in fact get a splitting of the homology groups.
