From Munkres Topology section $58$:
Why is a deformation retraction defined as $H: X \times I \to X$ instead of $H: X \times I \to A$?
It seems it would make more sense to define it as $H: X \times I \to A$ and get rid of the composition $j \circ r$.
Is this not possible?
On
Maybe it makes more sense to define before what a retraction is:
Definition. Let $X$ be a topological space and $A$ be a subset of $X$. A continuous map $r\colon X\rightarrow A$ such that $r_{\vert A}=\textrm{id}_A$ is called a retraction of $X$ on $A$. In other word, $r\circ i=\textrm{id}_A$, where $i:A\hookrightarrow X$ stands for the inclusion of $A$ in $X$.
Example. The map $\displaystyle r\colon\mathbb{C}^*\rightarrow\mathbb{S}^1,z\mapsto\frac{z}{|z|}$ is a retraction.
Using this definition, one has that:
Proposition. Let $X$ be a topological space and $A$ be a subset of $X$, then $X$ is a deformation retraction of $X$ if and only if there exists an homotopy between a retraction of $X$ on $A$ and the identity of $X$.
Exemple. The map $\displaystyle H\colon\mathbb{C}^*\times[0,1]\rightarrow\mathbb{C}^*,(z,t)\mapsto (1-t)z+t\frac{z}{|z|}$ is a deformation retraction. Notice that the image of $H$ is not included in $\mathbb{S}^1$, so according to your definition, it would not be a deformation retraction.
The heuristic for $X$ being a retraction of $A$ is that one can continuously squeeze/pull $X$, that is without tearing it apart, onto $A$ and this with $A$ being steady during the transformation. Furthermore, $X$ is a deformation retract of $A$ if this squeezing operation can be done gradually, that is $X$ being continuously squeezed onto $A$. Namely, $X$ is step by step shifted from its original appearance to his squeeze state.
With the definition you proposed, I cannot really grab any heuristic. It doesn't seem to make that much sense to ask for all points of $X$ being in $A$ from the beginning of the deformation.
Since part of the definition is that $H(x,0)=x$ for all $x\in X$, every $x\in X$ must be in the codomain of $H$. You only know that $H(x,t)\in A$ when $t=1$, and for other values of $t$ this need not be true (and in fact it must fail to be true if $t=0$ and $x\not\in A$).