In the long exact sequence $\cdots \to \tilde{H} (A) \to \tilde{H}(X) \to H(X, A) \to \cdots$, why isn’t the third homology in the sequence given in a reduced homology?
Also I am wondering why the order is specifically $A\to X\to (X, A)$. Is there any intuitive motivation behind it?
The answer to your second question is an algebraic one. Let $X$ be a top. space and $A\subset X$ a subspace. Let $(C_\bullet(X),\partial_\bullet)$ be a chain complex. The relative chaingroup $C_n(X,A)$ is defined as the quotient $$ C_n(X,A):= C_n(X)\big/C_n(A)$$
Observe that the differential $\partial_\bullet$ takes $$C_\bullet(A)\to C_{\bullet-1}(A)$$ which then induces a complex of quotients $$\cdots \to C_n(X,A)\to C_{n-1}(X,A)\to C_{n-2}(X,A) \to \cdots$$
Which gives rise to a short exact sequence $$0\to C_n(A)\xrightarrow{i} C_n(X)\xrightarrow{p} C_n(X,A) \to 0 \quad (*)$$ where $i$ is the inclusion and $p$ the quotient map. Since the differential $\partial_\bullet$ is part of each of the chain complexes $$C_\bullet(A), C_\bullet(X), C_\bullet(X,A)$$ we can interpret $(*)$ as a short exact sequence of chain complexes for which you'll get the homology groups $H_n(-)$ of the respective complexes $C_\bullet(A), C_\bullet(X), C_\bullet(X,A)$ by the snake lemma.
The snake lemma provides the long exact homology sequence you are looking at $$\cdots H_{n}(A) \to H_{n}(X) \to H_{n}(X,A) \xrightarrow{\rho} H_{n-1}(A)\to \cdots$$ where $\rho$ denotes the connecting map.
So generally speaking, the sequence $$H_n(A)\to H_n(X)\to H_n(X,A)$$ comes from the short exact sequence $(*)$ and applying the snake lemma to get the long exact sequence of the respective homology groups.