Let's $\Delta$ be a triangle in $X$, and $\Delta'$ a comparison triangle for $\Delta$ in Euclidean space with distance $d'$. For a point $x$ on the boundary of $\Delta$, we will denote by $x'$ its image on $\Delta'$ (if $x$ is on a side $[p,q]$ of $\Delta$, then $x'$ is such that $d(x,p) = d'(x',p')$ and $d(x,q) = d'(x',q')$)
Condition $(ii)$ says that the distance $d(p,m)$ between $p$ and the middle point $m$ of $[q,r]$ must be less than the distance $d'(p',m')$ between $p'$ and $m'$. Indeed, because geodesics are unique in CAT(0) spaces, the point $m$ is unique and thus is a middle point between $q$ and $r$. Moreover in Euclidean space, as you noticed, we have that $d'(p',q') + d'(p',r') = 2d'(p',m') + 2d'(q',r')$ for any triangle $p'q'r'$ and $m'$ middle point of $[q',r']$.
Condition $(i)$ says that the distance $d(p,x)$ between $p$ and any point $x$ on $[q,r]$ must be less than $d'(p',x')$.
So $(i)$ implies $(ii)$ by taking $x = m$.
To prove $(i)$ from $(ii)$, apply condition $(ii)$ on the triangles with vertices $\{p,q,m\}$ and $\{p,r,m\}$. For $x'$ the middle point of $[q,m]$ and $y'$ the middle point of $[m,r]$, you get that $d'(p',x') \leq d(p,x)$ and $d'(p',y') \leq d(p,y)$. More generally, if $x$ and $y$ are two points on $[q,r]$ for which $d'(p',x') \leq d(p,x)$ and $d'(p',y') \leq d(p,y)$ and $w$ is the middle point of $[x',y']$, then $d'(p',w') \leq d(p,w)$. Let $M$ be the closure of $\{q,r\}$ under taking middle points. Thus we have that for any $w$ in $M$, $d'(p',w') \leq d(p,w)$. Now any point $x \in [q,r]$ is the limit of a sequence of points in $M$, so by continuity $d(p',x') \leq d(p,x)$.