I'll discuss separately the question you asked and the exercise in the notes you linked to.
The question you asked is trivial as long as $k\geq3$. Just let $f$ be a projection onto the $3$-dimensional subspace spanned by $x,y,z$. You don't even need $\epsilon$, since $x,y,z$, and the angle are preserved exactly. (In fact, even if $k=2$, you can preserve the angle exactly by projecting orthogonally to the subspace spanned by $x-y$ and $z-y$.)
Of course, this isn't what the exercise (or the Johnson-Lindenstrauss lemma) is about. You wanted to approximately preserve one angle. The analog for angles of Johnson-Lindenstrauss wants to approximately preserve all angles between points from some given $q$-element set. To do this, one needs the target space to have dimension of order $(\log q)/(\epsilon^2)$; note the log of the number $q$ of points, not of the dimension of the space in which they lie.
Once the statement of the problem has been corrected to obtain for angles what Johnson-Lindenstrauss gives for distances, I want to use Paul Siegel's comment, that this follows from continuity of arccos. The only catch is that, at the ends $\pm1$ of its domain, arccos, though continuous, isn't differentiable. In other words, when $x,y,z$ are collinear, one can't trivially estimate the error in $\theta$ in terms of the errors in the distances between $x,y,z$. But one can estimate it with a bit of work. I'm too tired to attempt that work now, but it looks to me as if we might need $\epsilon^4$ rather than $\epsilon^2$ in the estimate for the dimension of the target space. (If we're guaranteed that no three of the $q$ points are collinear, then $\epsilon^2$ should suffice but the implicit constant in "of order $(\log q)/(\epsilon^2)$" will depend on how far from collinear the points are.)
Technicality: Worse than collinearity would be coincidence. If $y$ coincides with $x$ or $z$, then the angle at $y$ is undefined and the question disappears.