These confusion arise because we separate spacetime in space +time. We should think that spacetime is a four dimensional manifold
In Special relativity spacetime is the tuple $(\mathcal{R}^4, \mathcal{O}, \mathcal{A}^{\uparrow}, \nabla, \eta, T,)$,
where $\nabla$ is a torsion free connection $\eta=(+---)$ a Lorentzian metric,$\mathcal{O}$ the standard topology,$\mathcal{A}^{\uparrow}$ an atlas and $T$ a time-orientation.
A $\textbf{clock}$ carried by a specific observer $(\gamma, e)$ will measure a $\textbf{time}$
$$
time := \int_{\tau_0}^{\tau_1} d\tau \sqrt{ \eta_{\lambda(\tau)}(v_{\lambda,\lambda(\tau)}, v_{\lambda,\lambda(\tau)}) }
\tag1$$
between the two $''\underline{events}''$
$$
\lambda(\tau_0) \quad \quad \quad \, \text{ "start the clock'' }
$$
and
$$
\lambda(\tau_1) \quad \quad \quad \, \text{ "stop the clock'' }
$$
Here $v_{\lambda,\lambda(\tau)})$ is the tangent vector of the the curve $\lambda$ at the point $\lambda(\tau)$
Lets choose coordinates $(x_0=ct,x_1=x,x_2=y,x_3=z)$ such that the trajectory of observer $A$ in spacetime is given by
$$\lambda_A(\tau)=(c\tau,0,0,0)$$
and the trajectory of observer $B$ in spacetime is given by $$\lambda_B=(c\tau,d-v\tau,0,0)$$From these equation we see that $A$ and $B$ meet at $\tau= \frac{d}{v}$
In these coordinates the metric is given by $$\eta=dt\otimes dt-dx\otimes dx-dy\otimes dy-dz\otimes dz \tag 2$$
and the tangents vectors are given by
$$v^A_{\lambda,\lambda(\tau)}=c \frac{\partial}{\partial t} \tag 3$$
$$v^B_{\lambda,\lambda(\tau)}=c \frac{\partial}{\partial t}-v\frac{\partial}{\partial t} \tag 4$$
so we have
$$n(v^A_{\lambda,\lambda(\tau)},v^A_{\lambda,\lambda(\tau)})=c^2 \tag2$$
$$n(v^B_{\lambda,\lambda(\tau)},v^B_{\lambda,\lambda(\tau)})=c^2-v^2 \tag3$$
than the time elapsed until their meeting for observer $A$ is
$$t_A=\frac{1}{c}\int_0^\frac{d}{v}\sqrt{c^2}=\frac{d}{v}$$
The time for $B$ is given by
$$t_B=\frac{1}{c}\int_0^\frac{d}{v}\sqrt{c^2-v^2}=\frac{1}{\gamma}\frac{d}{v}$$
Eq.1 is coordinate independent so if we make the calculation from the perspective of $B$ we should obtain the same result.
