A property of $\gamma_1$ extensions of real closed fields. For each real closed ordered field $\mathbb{F}$, $\sum_\omega({}^m\mathbb{F},S)$ denotes the set of all definable subsets in ${}^m\mathbb{F}$ with parameters in S.
We consider $\sum_\omega({}^m\mathbb{F},S)$ as a boolean algebra.
Then, for each real closed ordered fields $\mathbb{H_0} \subset \mathbb{H_1} $, and for each $\langle a_1,...,a_m\rangle \in {}^m\mathbb{H}_1$, $\{A \cap {}^m\mathbb{H_0} \mid \langle a_1 ,...,a_m\rangle \in A \in \sum_\omega({}^m\mathbb{H_1}, \mathbb{H_0})\}$ is an ultrafilter over $\sum_\omega({}^m\mathbb{H_0}, \mathbb{H_0})$.
We say $\mathbb{H}_1$ is $\gamma_1$ over $\mathbb{H}_0$ if $\{A \cap {}^m\mathbb{H_0} \mid \langle a_1 ,...,a_m\rangle \in A \in \sum_\omega({}^m\mathbb{H_1}, \mathbb{H_0})\}$ is a countably generated ultrafilter over $\sum_\omega({}^m\mathbb{H_0}, \mathbb{H_0})$ for each finite sequence $\langle a_1,...,a_m\rangle$ in $\mathbb{H}_1$.
If $\mathbb{H}_1$ is $\gamma_1$ over $\mathbb{H}_0$ and if $\mathbb{H}_2$ is $\gamma_1$ over $\mathbb{H}_1$, then $\mathbb{H}_2$ is $\gamma_1$ over $\mathbb{H}_0$ ?
 A: I'll give you a purely model-theoretic proof, which has nothing to do with real closed fields. Let $T$ be a complete theory, and let $M\preceq N$ be an elementary extension of models of $T$. We say that $N$ is countably determined over $M$ if for every tuple $a\in N^n$, there is a countable set of formulas $\Sigma(x)$ with parameters from $M$ such that $\Sigma(x)\cup \text{EDiag}(M)$ is consistent, and $\Sigma(x)\cup \text{EDiag}(M)\models \text{tp}(a/M)$, where $\text{EDiag}(M)$ is the elementary diagram of $M$. This is the notion that you call $\gamma_1$ in the context of real closed fields. 
For context, we would say that $N$ is atomic over $M$ if we make the stronger requirement that $\Sigma(x)$ is finite (equivalently a single formula, since we can replace a finite set of formulas with their conjunction). The proof below is really just a generalization of the standard argument for the same claim, with "atomic" in place of "countably determined". The main idea is exactly the same. 
Claim: Let $M_0\preceq M_1\preceq M_2$. If $M_1$ is countably determined over $M_0$ and $M_2$ is countably determined over $M_1$, then $M_2$ is countably determined over $M_0$. 
Proof: Let's agree to work in a language with constant symbols naming every element of $M_0$. The result is that we never need to mention parameters from $M_0$: when I explicitly name parameters in formulas, they will always come from $M_1\setminus M_0$. 
Let $a\in M_2^n$ be a tuple from $M_2$. Since $M_2$ is countably determined over $M_1$, there is a countable set of formulas $\Sigma(x)$ with parameters from $M_1$ such that $\Sigma(x)\cup \text{EDiag}(M_1)\models \text{tp}(a/M_1)$ and is consistent. We may assume that $\Sigma(x)$ is closed under finite conjunctions. 
Let $\varphi(x,b)$ be a formula in $\Sigma(x)$. By our convention above, $b$ is a tuple of parameters from $M_1\setminus M_0$. Now $M_1$ is countably determined over $M_0$, so there is a countable set of formulas $\Delta_b(y)$ with parameters from $M_0$ such that $\Delta_b(y)\cup \text{EDiag}(M_0)\models \text{tp}(b/M_0)$ and is consistent. We may also assume that $\Delta_b(y)$ is closed under finite conjunctions. 
We define a new set of formulas $\Sigma^*(x)$ with parameters from $M_0$ as follows: $$\Sigma^*(x) = \bigcup_{\varphi(x,b)\in \Sigma(x)} \{\exists y\, (\varphi(x,y)\land \psi(y))\mid \psi(y)\in \Delta_b(y)\}.$$
Note that $\Sigma^*(x)$ is countable, since it is a countable union of countable sets of formulas. We will show that $\Sigma^*(x)\cup \text{EDiag}(M_0)\models \text{tp}(a/M_0)$. 
So let $\theta(x)$ be any formula in $\text{tp}(a/M_0)$. In particular, $\theta(x)\in \text{tp}(a/M_1)$, so $\Sigma(x)\cup \text{EDiag}(M_1)\models \theta(x)$. By compactness, and since $\Sigma(x)$ and $\text{EDiag}(M_1)$ are closed under finite conjunctions, there is a formula $\varphi(x,b)\in \Sigma(x)$ and a formula $\chi(b,d)\in \text{EDiag}(M_1)$ such that $\varphi(x,b)\land \chi(b,d)\models \theta(x)$. Note that we may assume that $\chi$ includes the tuple of parameters $b$. By our convention, $d$ is the tuple of additional parameters from $M_1\setminus M_0$ used in the formula. 
In particular, the elements of $d$ are not included in the tuple $b$ or mentioned in $\theta$, so $\varphi(x,b)\land (\exists w\, \chi(b,w))\models \theta(x)$. And $b$ is not mentioned in $\theta$, so $\exists y\, (\varphi(x,y)\land (\exists w\, \chi(y,w))\models \theta(x)$.
Next, observe that $\exists w\, \chi(y,w)\in \text{tp}(b/M_0)$, so $\Delta_b(y)\cup \text{Ediag}(M_0)\models \exists w\, \chi(y,w)$. By compactness, and since $\Delta_b(y)$ and $\text{EDiag}(M_0)$ are closed under finite conjunctions, there is a formula $\psi(y)\in \Delta_b(y)$ and a formula $\delta\in \text{EDiag}(M_0)$ such that $\psi(y)\land \delta \models \exists w\, \chi(y,w)$.
Putting all of the above together, we're done: From $\Sigma^*(x) \cup \text{Ediag}(M_0)$, we have $\exists y\, (\varphi(x,y)\land \psi(y))$ and $\delta$ as assumptions. Together, these entail $\exists y\, (\varphi(x,y)\land \psi(y)\land \delta)$. But since $\psi(y)\land \delta \models \exists w\, \chi(y,w)$, this entails $\exists y\, (\varphi(x,y)\land \exists w\, \chi(y,w))$, which entails $\theta(x)$ as observed above. 
