Let $i: R \subset A$ be an integral extension of discrete valuation rings and denote by $$f: \operatorname{Spec}(A) \to \operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $\operatorname{Spec}(R) = \{\sigma_R, \eta_R\}$ (resp. $\operatorname{Spec}(A) = \{\sigma_A, \eta_A\}$ where $\sigma$ is the unique closed and $\eta$ the unique generic points)
denote
$K_R := \kappa(\eta_R) = \operatorname{Frac}(R)$
$K_A := \kappa(\eta_A) = \operatorname{Frac}(A)$
and
$k_R := \kappa(\sigma_R)= \mathcal{O}_{R, \sigma_R}/m_{\sigma_R}\mathcal{O}_{R, \sigma_R}$
$ k_A := \kappa(\sigma_A)$
we obtain field extensions $K_R\subset K_A$ and $k_R\subset k_A$.
I want to show that:
$f$ is a unramified morphism (in sense of scheme morphism) $\Leftrightarrow$
$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.
Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.
My attempts:
Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:
So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.
The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{\sigma_R}$ generates $m_{\sigma_A}$ ($m_{\eta_R}$and $m_{\eta_A}=0$ so it's ok)
The author's hint was to use exersise 3.1.8 (page 90):
So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?
Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?
Hint: You can show that $A/R$ is in fact finite free with $\mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.