It suffices to show the polyonomial $n(x+m^2)(x-k_1)\cdots(x-k_{p-2})-2$ is irreducible over $\mathbb{Q}$, notice that this polynomial is primitive. So it suffices to show that it is irreducible over $\mathbb{Z}$, this can be proved by Eisenstein criterion when modulo 2.
The above arguement only uses the assumption that $m,k_i$ are even and $n$ is odd.
If $h$ has no multiple roots, then when we draw the picture of $f$, it is easy to see that all roots of $f$ are real and distinct.
E.g. $h=(x+2^2)(x-4)$ and $2/n=2$, then $f=x^2-18$ has two different real roots.
Edit: Our assumptions are that $k_i$ are distinct and $p$ is odd and $k_i>0$, $m>0$. The arguments for irreducibility still work. Now denote $h_2=x^2+m$ and $h_1=(x-k_1)(x-k_2)(x-k_{p-2})$. Consider the equation $h_1=s/h_2$, where $s=2/n, draw the picture to see that $h_1$ and $s/h_2$ intersects exactly $p-2$ different real points. We can use intermediate value theorem(continuity) to make this rigorous.