Suppose that such a function $f$ exists. Injectivity of $g$ implies the injectivity of the continuous function $f$, which in turn is strictly monotone. As $g$ is increasing we conclude that $f$ is increasing and $k$ is an odd number. Moreover, $f$ is not surjective.

Denote by $f^{[k]} = f \circ f \circ \dots \circ f$ ($k$ times $f$). Because $f(\mathbb{R})$ is an interval with $(-\infty, 0) = g(\mathbb{R}) = f(f^{k-1}(\mathbb{R})) \subset f(\mathbb{R})$, we deduce $f$ is bounded from above.

Let $m \in \mathbb{R}$ be such that $f(x) < m, \forall x \in \mathbb{R}$. Then $f^{[k-1]}(x) < m$, for all $x \in \mathbb{R}$, so $g(x) = f(f^{[k-1]}(x)) > f(m)$, for all $x \in \mathbb{R}$, in contradiction with $g(\mathbb{R}) = (-\infty, 0)$.