Finding a composite function $h$ from $g$ and $f$ defined by $h= g \circ f$ is much easier than decompose a function $h$ into its generating functions when only one of them is known.
Given a composite function $h = g \circ f$ and only one of its generating $f$, how can we find the second generating function $g$ such that $h = g \circ f$ ? If so, state your assumption if any and prove it. If not, why not? Is $g$ unique?
Now consider $h = g \circ f$ a given polynomial of degree $n$ with its leading coefficient $c_n$ of $x^n$, $c_n \neq 0$ and similarly $f$ be of the form $x+k$ for some known $k$. Can we find $g$? If so, how?
If $f$ has a left inverse $f_l$, that is $f_l \circ f = id$ we can pick $g=h \circ f_l$, $g \circ f (x) = h(f_l(f(x)))=h(x) $.
$g$ need not be unique, for example, let
$$h: \mathbb{R}^+ \rightarrow \mathbb{R}^+, h(x)=x,$$
$$f: \mathbb{R}^+ \rightarrow \mathbb{R}^+, f(x)=x$$
consider two different function
$$g_1:\mathbb{R}^+ \rightarrow \mathbb{R}^+, g_1(x)=x$$
$$g_2: \mathbb{R} \rightarrow \mathbb{R}^+, g_2(x) = \begin{cases}x, & x \in \mathbb{R}^+ \\ 0, & x \notin \mathbb{R}^+\end{cases}$$
We can see that $g_1 \circ f = g_2 \circ f$.
For the special case of $f(x)=x+k$, we can choose $g=h(x-k)$, resulting in
$$g \circ f (x) = g(f(x))= g(x+k)=h((x+k)-k)=h(x)$$