Direct sum of the group of $n$-chains $S_n(X)$ for all $n$ induces the direct sum of the chain complex $S_{\ast}(X)$? Let $(X,A)$ be a topological space pair in $(\mathbf{Top}^2)$, i.e $A$ is a subspace of $X$. For any $q \geq 0$, we have the decomposition
$$
S_q(X) = S_q(A) \oplus S_q(X,A). \quad \quad (\star)
$$

Reasons: Since any $q$-simplex $\sigma: \Delta_q \rightarrow A$ of $A$ is a $q$-simplex of $X$, and hence the basis of $S_q(A)$ (which is exactly the $q$-simplexes of $A$) is a subset of the basis of $S_q(X)$ (which consists all $q$-simplexes of $X$). So $S_q(A)$ is a direct summand of $S_q(X)$, and hence we obtain $(\star)$.

My main question: Does $(\star)$ means that we have the following decomposition $(\star\star)$ of chain complexes?
$$
S_{\ast}(X) = S_{\ast}(A) \oplus S_{\ast}(X,A) \quad \quad (\star\star)
$$
I shall explain my question further as follows:
1. My attempts - a counterexample: This should be incorrect, since if $(\star\star)$ holds, we shall apply the homology functor $H_{\ast}$ on both sides and obtain
$$
H_{\ast}(X) = H_{\ast}(A) \oplus H_{\ast}(X,A) \quad\quad(\star\star\star)
$$
However, let $X=\mathbb{D}^n$,$A=\mathbb{S}^{n-1}$ and $\ast = n$, then
$$
\mathrm{LHS} = H_{n}(X) = H_{n}(\mathbb{D}^n) = 0,
$$
while
$$
\mathrm{RHS} = H_{n}(A) \oplus H_{n}(X,A) = H_{n}(\mathbb{S}^{n-1}) \oplus H_{n}(\mathbb{D}^n, \mathbb{S}^{n-1}) = \mathbb{Z},
$$
which yields $\mathrm{LHS} \neq \mathrm{RHS}$.
2. My questions: From the counterexample above, $(\star\star)$ should be incorrect, but I don't know why $(\star)$ DOES NOT imply $(\star\star)$? OR is my argument in "Reasons" incorrect, which means that even $(\star)$ does not holds in general?
3. Wild guess: Does these things related to the contractiablity of $A$ as a subspace of $X$. Since once $A$ is a retract of $X$, we shall obtain $r: X \rightarrow A$, satisfying
$$\iota \circ r = \mathrm{id}_{A} \quad\quad (\ast)$$
where $\iota: A \rightarrow X$ is the embedding. Applying the functor $S_{\ast}$ on both sides of $(\ast)$, we see that the exact sequence of chain complexes
$$
0 \rightarrow S_{\ast}(A) \rightarrow S_{\ast}(X) \rightarrow S_{\ast}(X,A) \rightarrow 0
$$
splits, and hence we obtain $(\star\star)$ again, then we get $(\star\star\star)$. So the counterexample shows that $\mathbb{S}^{n-1}$ is not a retract of $\mathbb{D}^n$.
4. Further question: Is my wild guess in "3" correct?
Sorry for such a long post and thank you all for commenting and answering!
 A: The short answer to your main question is "no". If $A_*$, $B_*$, and $C_*$ are chain complexes, having $C_q = A_q \oplus B_q$ for each $q$ in some sense doesn't guarantee that $C_* = A_* \oplus B_*$ as chain complexes. The relevant maps between $C_q$ and $A_q \oplus B_q$ need to also respect the boundary maps in the chain complexes in order for that to be true.
The counterexample you gave in "1" is essentially correct.
You know $(\star \star \star)$ doesn't hold, which leads you to conclude that $(\star \star)$ doesn't hold and the question you're asking in "2" is essentially "how come?". One way to arrive at an answer to that is to be more explicit about what's actually going on with $(\star \star)$, and $(\star)$ in your counterexample.
Concretize, concretize, concretize
For concreteness you might as well fix $n = 1$. So $X = [0,1]$, and $A = \{0, 1\}$. The relevant homology groups are trivial except $H_0(A) = \mathbb Z^2$, $H_0(X) = \mathbb Z$, and $H_1(X, A) = \mathbb Z$. In particular, we have $H_0(X) \neq H_0(A) \oplus H_0(X,A)$. What gives? Let's work out what these things should be from first principles.
You've got $H_0(A) = S_0(A) / \partial S_1(A).$ In a rare case of sanity, we literally just have $S_0(A) = \mathbb Z^2$ and $\partial S_1(A) = 0$, so we can be really concrete about this.
You've got $H_0(X) = S_0(X) / \partial S_1(X).$ We have $S_0(X)$ is the free abelian group generated by points in $X$, and $\partial S_1(X)$ is generated by differences between arbitrary pairs of those generators. This is messy in that the two spaces are infinite-dimensional, but still pretty concrete, and it's hopefully clear why the quotient is indeed just $\mathbb Z$.
Now what about $H_0(X,A)$, which is $S_0(X,A) / \partial S_1(X,A)$? We have conflicting information about this, according to one argument it's trivial, and according to another argument it should be $\mathbb Z$. The claim that it's trivial just means every element of $S_0(X,A)$ is a boundary of some element of $S_1(X,A)$. This is something that you can dig into "by hand". Here are the two arguments when you unwrap them.
(1) It's trivial. (You can get this by concretizing whatever abstract argument you have for this, e.g. the argument in terms of the long exact sequence.) We have $S_0(X,A) = S_0(X) / S_0(A)$, so an element of $S_0(X,A)$ is an abstract sum of points in X, except two abstract sums are considered the same if their difference is a sum of points in A. We have $S_1(X,A) = S_1(X) / S_1(A)$, so an element of $S_1(X, A)$ is an abstract sum of paths in $X$, except two abstract sums are considered the same if their difference is paths that live entirely in $A$. The boundary map between them is "the obvious thing" (take the signed boundary of each path), though it takes a bit of thought to see that it's well-defined. Is it true that every abstract sum of points in X (modulo points in A) is the boundary of some abstract sum of paths in X (modulo paths in A)? Yes. Given an abstract sum of points in X, pick a representative with total signed sum of points equal to 0 (since you're free to "add or remove points in A"). Then pair up the positive points with the negative points and draw a path for each pair. The abstract sum of these paths hits the element of $S_0(X, A)$ that you started with.
(2) It's not trivial. We have $S_0(X) = S_0(A) \oplus S_0(X, A)$. Since $S_0(X)$ is abstract sums of points in $X$, and $S_0(A)$ is abstract sums of points in $A$, we have that $S_0(X, A)$ is abstract sums of points in $X$ which aren't in $A$. Similarly, $S_1(X, A)$ is abstract sums of paths in $X$ which aren't (entirely) in $A$. Is it true that every element of $S_0(X, A)$ is a boundary of some element of $S_1(X, A)$? No, the only way an abstract sum of points can be a boundary of an abstract sum of paths is if the total signed sum of all the points is 0.
So which of these is right? Is the boundary map $\partial: S_1(X, A) \to S_0(X, A)$ surjective or isn't it? Even more concretely, is there an element of $S_1(X, A)$ which maps, via the boundary map, to the element of $S_0(X, A)$ corresponding to just the midpoint $1/2 \in X$? Argument (1) constructs one, e.g. a path in $X$ from $0 \in A \subseteq X$ to $1/2 \in X$ maps to that element of $S_0(X, A)$ according to argument (1). But not according to argument (2).
Well what does that path map to according to argument (2)? It should map to the abstract sum of positive copy of $1/2 \in X$ and one negative copy of $0 \in X$. Oops, according to argument (2), that's not even an element of $S_0(X, A)$ since it uses points in $A$! So if we believe argument (2) then we've run into worse issues; the boundary map from $S_1(X, A)$ to $S_0(X, A)$ doesn't even make sense!
Then abstract
Hopefully that's enough exposition to show that just by concretizing things as much as possible, you're forced to realize that there's something pretty sketchy about argument (2), and that this element of $S_1(X, A)$ really should have some image with the boundary map.
One way to phrase this is that your mistake was to implicitly treat $S_q(X, A)$ as contained in $S_q(X)$ when in fact it is a quotient of $S_q(X)$. You've perhaps hidden more than you realized in the claim $S_q(X) = S_q(A) \oplus S_q(A,X).$
Specifically, it's true that for each $q$ the exact sequence $$0 \to S_q(A) \to S_q(X) \to S_q(A,X) \to 0$$ splits, but the way you phrased it brushes the role of the splitting map $S_q(A, X) \to S_q(X)$ under the carpet. It turns out the splitting map you have in mind does not play well with boundaries, so it doesn't correspond to a chain complex map, which means it doesn't give you a split exact sequence of chain complexes $$0 \to S_*(A) \to S_*(X) \to S_*(A,X) \to 0.$$
