Let $z_1, \text{ } z_2 \in \mathbb{C}$. We now that:
$$\left| z_1 + z_2 \right| = \left| z_1 \right| + \left| z_2 \right| \iff \frac{z_1}{z_2} > 0 \vee \left( z_1 = 0 \vee z_2 = 0 \right)$$
My question is: can we extend this to elements of $\mathbb{H}$:
$$\left| q_1 + q_2 \right| = \left| q_1 \right| + \left| q_2 \right| \iff \frac{q_1}{q_2} > 0 \vee \left( q_1 = 0 \vee q_2 = 0 \right), \text{ } q_1, \text{ } q_2 \in \mathbb{H}$$
In any inner product space $V$, we have $(*)$ equality holds in the triangle inequality for vectors $z_1$ and $z_2$ iff there is a ray $r$ from $0$ through $z_0$ such that $z_2$ lies on $r$. To prove $(*)$ it suffices to assume $\dim(V) \le 2$, since vectors outside the span of $z_1$ and $z_2$ are not relevant.
For $V = \Bbb{C}$, $(*)$ is what your first equivalence states using the idiom $\frac{z_1}{z_2} > 0$ to capture the condition that $z_2$ lies on the ray from $0$ through $z_1$ when $z_1$ and $z_2$ are both non-zero (via the convention that $z > 0$ is false if $z$ is not real). So the proof of your first equivalence proves $(*)$
For $V = \Bbb{H}$, your second equivalence is also valid using $(*)$ as it amounts to the statement that $z_2$ lies on the ray through $z_1$ from $0$.