Are $\pm f\sqrt{1+g^2}$ and $\pm fg\sqrt{1+g^2}$ smooth if $f,fg,fg^2$ are smooth?

Still no. Consider the function $f(x)=(x^2+r^2)^2$, $g(x)=\frac x{x^2+r^2}$ with small $r>0$. Then $f$, $fg$, $fg^2$ are polynomials of degree $\le 4$ with bounded coefficients but $f\sqrt{1+g^2}$ is very close to $x^2|x|\sqrt{1+x^2}$ as close to $0$ as you wish when $r$ is small enough, so the maximum of the fourth derivative in an arbitrarily small neighborhood of the origin can be forced to be very large by choosing $r$ small enough. Now just take your favorite $C^\infty$ function $\psi$ that is $1$ on $[-1,1]$ and is supported on $[-2,2]$ and use $\psi f$ and $\psi g$ instead of $f$ and $g$. You'll get a compactly supported building block that you can scale and translate with the possibility to blow up the fourth derivative by choosing $r$ last. So just scale to some disjoint intervals $I_j$ accumulating to $0$ with sufficiently fast decaying heights to make individual multiplications by controlled polynomials irrelevant after which choose $r_j$. Near the center of each interval $I_j$ the function $f$ is strictly positive, so $h$ is of no use there. What may really help (no guarantee though) is to assume that $g$ is continuous, but that is, probably, too much for your purposes.

$\newcommand{\de}{\delta}$This is to provide a detalization/formalization of fedja's answer.

For $r\in(0,1]$ and real $x$, let \begin{equation*} f_{0,r}(x):=(x^2+r^2)^2,\quad g_{0,r}(x):=\frac x{x^2+r^2}, \end{equation*} \begin{equation*} H_{0,r}:=f_{0,r}\sqrt{1+g_{0,r}^2},\quad F_{0,m,r}:=f_{0,r}g_{0,r}^m, \end{equation*} \begin{equation*} f_r:=f_{0,r}\psi,\quad g_r:=g_{0,r}\psi, \end{equation*} \begin{equation*} H_r:=f_r\sqrt{1+g_r^2},\quad F_{m,r}:=f_r g_r^m, \end{equation*} where $m\in\{0,1,2\}$ and $\psi$ is any function in $C^\infty(\mathbb R)$ such that $\psi=1$ on the interval $[-1/2,1/2]$ and $0$ outside the interval $[-1,1]$.

Then for each $k\in\{0,1,\dots\}$ we have $\max_{m=0}^2\sup_{0<r\le1}\|F_{0,m,r}^{(k)}\|_\infty<\infty$ and hence \begin{equation*} C_k:=\max_{m=0}^2\sup_{0<r\le1}\|F_{m,r}^{(k)}\|_\infty<\infty. \tag{1} \end{equation*} However, \begin{equation*} H_r^{(4)}(0)=H_{0,r}^{(4)}(0)=24-3/r^4\sim-3/r^4\to-\infty \end{equation*} as $r\downarrow0$; this crucial fact can be verified either by a direct calculation or by using (with $r$ fixed) the Maclaurin expansions $\sqrt{1+v}=1+v/2-v^2/8+o(v^2)$ (with $v=\dfrac u{(r^2+u)^2}$) and then $\dfrac1{(r^2+u)^2}=\dfrac1{r^4}\,\Big(1-\dfrac{2u}{r^2}\Big)+o(u^2)$ (with $u=x^2$).

For real $x$ and $m\in\{0,1,2\}$, let now \begin{equation*} f(x):=\sum_{j=1}^\infty a_j f_{r_j}\Big(\frac{x-c_j}{\de_j}\Big), \quad g(x):=\sum_{j=1}^\infty a_j g_{r_j}\Big(\frac{x-c_j}{\de_j}\Big), \end{equation*} \begin{equation*} F_m(x):=f(x)g(x)^m=\sum_{j=1}^\infty a_j F_{m,r_j}\Big(\frac{x-c_j}{\de_j}\Big) \end{equation*} \begin{equation*} H(x):=f(x)\sqrt{1+g(x)^2}=\sum_{j=1}^\infty a_j H_{r_j}\Big(\frac{x-c_j}{\de_j}\Big), \end{equation*} where \begin{equation*} r_j:=a_j:=e^{-j}, \quad c_j:=\tfrac12\,(x_j+x_{j+1}), \quad x_j:=1/j,\quad \de_j:=\tfrac12\,(x_j-x_{j+1})\sim1/(2j^2) \end{equation*} as $j\to\infty$.

Then, by (1) and dominated convergence, the functions $f=F_0$, $fg=F_1$, $fg^2=F_2$ are in $C^\infty(\mathbb R)$, with \begin{equation} \|F_m^{(k)}\|_\infty\le\sum_{j=1}^\infty a_j C_k/\de_j^k<\infty \end{equation} for all $k\in\{0,1,\dots\}$.

However, $hH\notin C^\infty(\mathbb R)$, and even $hH\notin C^4(\mathbb R)$, for any function $h\colon\mathbb R\to\{-1,1\}$. Indeed, assume the contrary. Then, for each natural $j$ and all $x\in[c_j-\de_j/2,c_j+\de_j/2]$, we have $h(x)H(x)=a_j h(x)H_{r_j}\big(\frac{x-c_j}{\de_j}\big)$, which will be continuous in $x$ at $x=c_j$ only if $h$ is the constant $1$ or the constant $-1$ in some neighborhood of $c_j$. Hence, \begin{equation} |(hH)^{(4)}(c_j)|=a_j|H^{(4)}(0)|/\de_j^4\sim 3a_j/(r_j^4\de_j^4)\to\infty \end{equation} as $j\to\infty$. Since $c_j\to0$ as $j\to\infty$, we see that $(hH)^{(4)}$ is unbounded in any neighborhood of $0$. So, $hH\notin C^4(\mathbb R)$, as claimed.