$\newcommand{\mc}{\mathcal}$The confusion seems to be over the following claim:
Claim: Let $\mc U$ be a countably incomplete ultrafilter, and let $E$ be a Banach space. If $(E)_{\mc U}$ is reflexive, then $E$ is super-reflexive.
To start with, we use the Eberlein-Smulian Theorem to observe that a Banach space $F$ is reflexive if and only if every separable subspace of $F$ is reflexive. [Question: Do we need Eberlein-Smulian to show this?]
Recall also that $\mc U$ being countably incomplete means that there is a nested sequence of sets $A_1 \supseteq A_2 \supseteq \cdots$ in $\mc U$ with $\cap_i A_i = \emptyset$. I think of this property as allowing us to embed sequential convergence into convergence along $\mc U$.
Finally, let us recall Theorem 6.3 in Heinrich's paper:
If $F$ is a separable Banach space finitely representable in $E$ then $F$ embeds isometrically into $(E)_{\mc U}$ for any countably incomplete $\mc U$.
Suppose towards a contradiction that $E$ is not super-reflexive, so there is a non-reflexive $F$ finitely representable in $E$. There is hence a separable subspace $F_0$ of $F$ which is not reflexive. Clearly $F_0$ is still finitely representably in $E$, and so isometric to a subspace of $(E)_{\mc U}$. Hence $(E)_{\mc U}$ is not reflexive, contrary to assumption.
We then proceed exactly as Jochen Glueck's comment.