I am slightly confused about the proof presented in Rudin. It says that the ordered field $Q$ is isomorphic to the ordered field $Q*$ whose elements are the rational cuts. It is this identification of $Q$ with $Q*$ which allows us to regard $Q$ as a subfield of $R$.
Now, I know that if I have to prove that $A$ is a subfield of $(B,+,.)$ then first I need to prove that $A$ is a subset of $B$ and the field operations in $B$ can be extended to $A$. For example, let $a, b \in A$ then $a+b \in A$ should be true.
Now for the above case, let $p$ and $q$ be two rational numbers each represented by the rational numbers less than them in $R$. Now if we add them using the $+$ operation in $R$, does that gives the rational number $p+q$ ? I don't think so.
We deal with your question about $p+q$. That is a large step toward proving that $\mathbb{Q}$, under the usual operations, and $\mathbb{Q}^\ast$, under the newly defined addition and multiplication, are isomorphic.
For any rational number $r$, let $r^\ast$ be the cut associated with $r$. So we want to prove that $(p+q)^\ast=p^\ast+q^\ast$.
Note that $(p+q)^\ast$ and $p^\ast +q^\ast$ are sets. Presumably you have already proved that in fact $p^\ast+q^\ast$ is a cut, since that is needed to show that the operation $+$ on cuts (the reals) is well-defined.
To show that two sets $X$ and $Y$ are equal, we show that any element of $X$ is an element of $Y$, and vice-versa.
So let $r\in (p+q)^\ast$. Then $r\lt p+q$. Let $t=\frac{p+q-r}{2}$, and let $x=p-t$, $y=q-t$. Note that $x+y=p+q-2t$ and therefore $x+y=r$.
Then $x\lt p$, so $x\in p^\ast$. Similarly, $y\in q^\ast$. So by definition $x+y\in p^\ast +q^\ast$. But $x+y=r$. This completes the proof in one direction.
The other direction is easier. Let $w\in p^\ast+q^\ast$. Then $w=x+y$ for some $x\in p^\ast$ and $y\in q^\ast$. So $x\lt p$ and $y\lt q$, and therefore $x+y\lt p+q$. It follows that $x+y\in (p+q)^\ast$, that is, $w\in (p+q)^\ast$.
There is more to do to show that the map that takes $r$ to $r^\ast$ is an isomorphism. We need to show that the mapping is onto (very easy), and one to one (straightforward). We also need to verify that multiplication also behaves well, that is, that $(pq)^\ast=p^\ast q^\ast$. One direction of this will take some work.