Another attempt at explaining why the first definition doesn't work, so assume that we are using that definition.
Let the only sort be $\mathsf{Exp}$, let $\mathcal{X}_{\mathsf{Exp}} = \{y,z\}$, and let $x$ be a variable (of sort $\mathsf{Exp}$) that is fresh for $\mathcal{X}$. Then $y,z \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}$ and $x \in \mathcal{B}[\mathcal{X},x]_{\mathsf{Exp}}$, which implies that
$$ \mathtt{let}(y; x.\!x) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}, $$
since $\mathtt{let}$ has generalised(generalised) arity $(\mathsf{Exp}, \mathsf{Exp}.\!\mathsf{Exp})\mathsf{Exp}$. In order to show (the false proposition) that
$$ \mathtt{let}(z; x.\!\mathtt{let}(y; x.\!x)) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}, $$
we would need $\mathtt{let}(y; x.\!x)$ to not only be in $\mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}$, but also in $\mathcal{B}[\mathcal{X,x}]_{\mathsf{Exp}}$. And this is what is impossible, since this requires $x$ to be fresh for $\mathcal{X},x$. (For ABTs with fresh renaming $\mathcal{X} \subseteq \mathcal{Y}$ implies $\mathcal{B}[\mathcal{X}] \subseteq \mathcal{B}[\mathcal{Y}]$, cf. Exercise 1.2, but this doesn't hold for ABTs without fresh renaming.)
Instead use the second definition of ABTs. Let $\rho(x) = x'$ be a fresh renaming of $x$ (relative to $\mathcal{X}$), and let $\rho'(x') = x''$ be a fresh renaming relative to $\mathcal{X},x$. Then $\hat{\rho}'(x') = \rho(x') = x'' \in \mathcal{B}[\mathcal{X},x',x'']_{\mathsf{Exp}}$, so we have
$$ \mathtt{let}(y; x'\!.\!x') \in \mathcal{B}[\mathcal{X},x']_{\mathsf{Exp}}, $$
since $\rho'$ was an arbitrary fresh renaming of $x'$. Next we have
$$ \hat{\rho}(\mathtt{let}(y; x.\!x)) = \mathtt{let}(y; x'\!.\!x') \in \mathcal{B}[\mathcal{X},x']_{\mathsf{Exp}}, $$
and since $\rho$ was an arbitrary fresh renaming of $x$, this implies that
$$ \mathtt{let}(z; x.\!\mathtt{let}(y; x.\!x)) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}. $$