No. For sufficiently nasty $f_i$'s, you can't even get a 3-element $z$ with the independence property that you specified. Choose, for each ordinal $\alpha<\omega_2$, a one-to-one map into $\omega_1$, and assemble all these maps into a single binary function $f$, so that, for each fixed $\alpha<\omega_2$, the function $\beta\mapsto f(\alpha,\beta)$ maps ordinals $\beta<\alpha$ one-to-one to countable ordinals. (I don't care what $f(\alpha,\beta)$ is when $\alpha\leq\beta$.) Similarly, let $g$ be a binary function such that, for each fixed $\alpha<\omega_1$, the function $\beta\mapsto g(\alpha,\beta)$ maps the ordinals $\beta<\alpha$ one-to-one to natural numbers. Now suppose $z$ is a 3-element set of ordinals, say $\xi<\eta<\theta$. Let $y\subseteq z$ consist of $\theta$ and whichever of $\xi$ and $\eta$ has the larger image under $f(\theta,-)$. Call that larger image $\rho$, and let $\sigma<\rho$ be the image of the other one of $\xi$ and $\eta$. Then the Skolem hull of $y$, in $(H_{\omega_2},\in,\lhd,f,g)$ contains all of the following: $\rho$ (because it's closed under $f$, which maps the pair of elements of $y$ to $\rho$), $g(\rho,\sigma)$ (because this is a natural number and therefore definable in $H_{\omega_2}$), $\sigma$ (because it's definable from $g(\rho,\sigma)$ and $\rho$, using $g$), and the element of $z-y$ (because it maps to $\sigma$ under $f(\theta,-)$ and is therefore definable from $\sigma$ and $\theta$ using $f$). So the Skolem hull of $y$, intersected with $z$, is not only $y$ but all of $z$.
