Since $T_1$ and $T_2$ are model complete, they are $\forall\exists$-axiomatizable, and $T_1\cup T_2$ is also $\forall\exists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1\cup T_2$.
So suppose $M\subseteq M'$, both models of $T_1\cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $\psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'\models \exists x\, \psi(x)$. You want to observe the following things:
It suffices to assume that $\psi(x)$ is a conjunction of atomic and negated atomic $(L_1\cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $\psi(x)$ is actually a conjunction $\varphi_1(x)\land \varphi_2(x)$, where $\varphi_1$ is an $L_1$-formula and $\varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $\psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $\psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'\models \psi(a')$ such that each element of $a'$ is in $M'\setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $\varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $\varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $\theta_1(y)$ and $\theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $\varphi_1$ and $\varphi_2$, respectively. Let $\theta'_1(y)$ be the conjunction of $\theta_1(y)$ and inequations $y_i\neq y_j$ for $i\neq j$, and similarly for $\theta'_2(z)$. Then look at the following sentence: $$\forall y\, \forall z\, ((\theta'_1(y)\land \theta'_2(z))\rightarrow \exists x\, (\varphi_1(x,y)\land \varphi_2(x,z))).$$
The model companion of $T_1\cup T_2$ is axiomatized by $T_1\cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1\cup T_2$. You also need to show the converse: that every existentially closed model of $T_1\cup T_2$ satisfies these extra axioms. Explicitly, given $\varphi_1(x,y)$ and $\varphi_2(x,z)$, if $M\models \theta_1'(b)\land \theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1\cup T_2$ such that $M'\models \varphi_1(a',b)\land \varphi_2(a',c)$ for some $a'\in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
Both theories are inconsistent.
For $T_1$, here's a hint: If $N\models T_1$, then $N$ is a model of $T$ in which $d$ is an $S$-maximal element. But a model of $T$ cannot have an $S$-maximal element. Why not?
For $T_2$, your answer is correct, but I would expect a more rigorous explanation. You wrote:
"if we applying the compatible theorem because finite groups of circles can scatter around the cartesian coordinate system, such that we won't be able to choose one that contained in all circles."
What is the "compatible theorem"? Do you mean the compactness theorem? What does "finite groups of circles can scatter around the cartesian coordinate system" mean precisely?
One way to give a very clear explanation is to find finitely many sentences in $T$ and finitely many of the new axioms $S(d,c_{a,b,r})$ in $T_2$ which are inconsistent. (Because of the compactness theorem, we know that we can always explain inconsistency of $T_2$ by finitely many axioms.) In fact, in this case a single sentence of $T$ and $2$ of the new axioms in $T_2$ will be enough.
It's surprising to me that your question asks about $$T_2=T\cup\{S(d,c_{a,b,r}):a,b\in \mathbb{R},r\in(0,\infty )\}$$ instead of $$T_2'=T\cup\{S(c_{a,b,r},d):a,b\in \mathbb{R},r\in(0,\infty )\}.$$ In fact, $T_2'$ is consistent. Can you prove this?
Taken together, the fact that $T_1$ is inconsistent but $T_2'$ is consistent makes a nice point: The theory $T$ rules out having a circle which contains all other circles, but it's possible to find a model with a circle $d$ which contains all of the standard circles (the elements named by constants in the standard model $M$).
Best Answer
First, the statement as written is incorrect: it's not an "if and only if". Completeness of $T_1\cap T_2$ implies consistency of $T_1\cup T_2$, but not the other way around.
As an aside, the nLab is a great resource for category theory, but it's probably not the best place to learn about logic (other than categorical logic).
#1: A theory is a set of sentences in some language $L$. We're looking at the theory $T_1\cup T_2$, which is a set of sentences in the language $(L_1\cup L_2)$. A theory is consistent if it does not prove $\bot$ (a contradiction). A theory is satisfiable if it has a model. By Gödel's Completeness Theorem, a first-order theory is consistent if and only if it is satisfiable, so these two terms are often used interchangeably.
#2: I prefer to say an $L$-theory $T$ is complete if $T\models \varphi$ or $T\models \lnot \varphi$ for every $L$-sentence $\varphi$ (i.e., $T$ entails $\varphi$ or its negation, not necessarily contains $\varphi$ or its negation). This is related to decision of whether or not to assume theories are always closed under entailment by convention.
#3: Here's how to derive Robinson Joint Consistency as a consequence of Craig Interpolation. Suppose $T_1\cap T_2$ is complete, and assume for contradiction that $T_1\cup T_2$ is inconsistent. By the compactness theorem, a finite subset of $T_1\cup T_2$ is inconsistent, say $\varphi_1,\dots,\varphi_n,\psi_1,\dots,\psi_m$, with all $\varphi_i\in T_1$ and all $\psi_j\in T_2$. Let $\varphi = \bigwedge_{i=1}^n \varphi_i$ and $\psi = \bigwedge_{j = 1}^m \psi_j$. Then $\varphi\land \psi$ is inconsistent, so $\varphi\models \lnot \psi$. By Craig Interpolation, there is some $(L_1\cap L_2)$-sentence $\theta$ such that $\varphi\models \theta$ and $\theta\models \lnot \psi$. Note that $\lnot\theta\models \lnot \varphi$.
Now since $(T_1\cap T_2)$ is complete, either $T_1\cap T_2\models \theta$ or $T_1\cap T_2\models \lnot\theta$. In the first case, $T_2\models\theta$, so $T_2\models \lnot\psi$, but $T_2\models \psi$, contradicting consistency of $T_2$. In the second case, $T_1\models \lnot\theta$, so $T_1\models \lnot \varphi$, but $T_1\models \varphi$, contradicting consistency of $T_1$.