Let $\tilde g(a,b)=(a,b)$ be the ``unnormalized'' version of $g$,
define $\tilde f(x):=\tilde g(h_1(x),h_2(x))$,
and note that $$
\frac12 L_D(\tilde f)
\le
L_D(f)
\le
L_D(\tilde f)
$$
where the inequality holds coordinate-wise;
this is because $\frac12\tilde g\le g\le\tilde g$.
Analogously,
$
\frac12 \hat L_S(\tilde f)
\le
\hat L_S(f)
\le
\hat L_S(\tilde f)
$.

Now let $m=\Theta(\frac{d}{\epsilon^2}\log\frac1\delta)$ be
large enough to guarantee that
$$ P\left\{ \sup_{h\in H} |L_D(h)-\hat L_S(h)|>\epsilon\right\}<\delta,$$
where $L_D(h),\hat L_S(h)$ are defined in the obvious way.

Thus, with probability $1-\delta$, each component of $L_D(\tilde f)$
is $\epsilon$-close to its empirical counterpart and their $\ell_1$-distance
is at most $2\epsilon$.

It follows that for $m$ as chosen above,
$$
P\left\{ \sup_{f\in F} ||L_D(f)-\hat L_S(f)|>2\epsilon\right\}<\delta.
$$