$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, we have 
$$\text{$f^{(k)}(a)=0$ if $h$ does not have all the derivatives at $a$.}\tag{1}$$ 

Indeed, take any $a\in\R$ and suppose that there is a sequence $(x_n)_{n\in\N}$ in $\R$ converging to $a$ such that for all $n$ we have $x_n\ne a$ and $f(x_n)=0$. This sequence may be assumed to be strictly monotone. So, by Rolle's theorem, for each $k\in\{0,1,\dots\}$ there is a sequence $(x_{k;n})_{n\in\N}$ such that $x_{k;n}\to a$ (as $n\to\infty$) and $f^{(k)}(x_{k;n})=0$ for all $n\in\N$, so that $f^{(k)}(a)=0$. 

So, if $f^{(k)}(a)\ne0$, then for all $x$ in a neighborhood $V$ of $a$ we have $f(x)\ne0$; hence, $h=(hf)/f$ is smooth on $V$ and therefore has all the derivatives at $a$. So, (1) is proved. 

Now it follows that $g_k:=f^{(k)}h$ must be smooth. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), for all $m\in\N$ we have $f^{(k)}(x)=o(|x-a|^m)$ as $x\to a$, so that $g_k(x)=o(|x-a|^m)$ as $x\to a$, so that $g_k^{(m)}(a)=0$, for all $m\in\N$.

Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. 
By (1), for all $m\in\N$ we have $f^{(k)}(x)=o(|x-a|^m)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|^m)$ as $x\to a$, and hence $g_{k,1}^{(m)}(a)=0$, for all $m\in\N$. 

However, $g_{0,2}$ is not continuous in general. E.g., let $f(x):=e^{-1/|x|}$ for $x\ne0$, with $f(0):=0$. Let 
$$\tilde h(x):=1(x>0)\int_0^x h_1(u)\,du,$$
where 
$$h_1(u):=\sum_{n\in\N}(-1)^n\,1(\tfrac1{n+1}<u<\tfrac1n).$$
Then $h_1$ is Lipschitz and $|\tilde h''(\tfrac1n)|=\infty$ for all $n\in\N$. Approximate $h_1$ closely enough by a Lipschitz function $h$ smooth enough on $(0,\infty)$ so that $|h''(\tfrac1n)|>e^{2n}$ for all $n$. Then $|g_{0,2}(\tfrac1n)|>e^{-n}e^{2n}\to\infty$ as $n\to\infty$. So, as claimed, $g_{0,2}$ is not continuous.