Invert an exponential and distinguish a composite from a square

For $y=e^{1-x^2}$, $\ln y=1-x^2$. Since $x>0$, choose $x=\sqrt{1-\ln y}$. Thus $f^{-1}:x\mapsto\sqrt{1-\ln x}$, $0<x<e$. For the composite, $g^2(x)=g(g(x))=\dfrac{1+(1+x)/x}{(1+x)/x}=\dfrac{2x+1}{x+1}$, $x>0$. This is a repeated application, not $[g(x)]^2$.

Write $g^2(x)=2-\dfrac1{x+1}$ for $x>0$. It increases from the excluded endpoint $(0,1)$ towards the horizontal asymptote $y=2$, so its range is $(1,2)$. For $m\geq0$, the line $y=mx+2$ is at least 2 on $x>0$, and cannot meet the curve. For $m<0$, the line-minus-curve difference tends to 1 as $x\to0^+$ and to $-\infty$ as $x\to\infty$, so it has an intersection. The required set is $\{m\in\mathbb R:m\geq0\}$.

The key rewrites $k(h(x))=2x^2-4x+15=2(x-1)^2+13=2[(x-1)^2+5]+3$, and gives $k:x\mapsto2x+3$, $x\in\mathbb R$. This is a valid function satisfying the equation. There is a qualification: $R_h=[5,\infty)$, so the equation forces $k(t)=2t+3$ only for $t\geq5$. With merely the printed domain $\mathbb R$, the values of $k(t)$ for $t<5$ are unrestricted. The printed global linear rule is one valid extension, not a uniquely determined answer without a further assumption.