Can the roots of spacetime interval be linearly added up? It seems ridiculous to add up roots of two intervals to get the total interval. Considering 3 events, $\Delta s_{13}$ shouldn't be equal to $\Delta s_{12}$+$\Delta s_{23}$. But if we consider the events in a moving particle's worldline, it seems reasonable to add up the proper times. What's wrong with the logic?
 A: The square root form for $\Delta s$ in terms of $v$ is correct for infinitesimal intervals, and really you should integrate it to find a finite spacetime interval:
$$\Delta s_{12}=\int_1^2 \sqrt{ds^2}.$$
So indeed if you consider three events,
$$\Delta s_{13}=\int_1^3 \sqrt{ds^2}=\int_1^2 \sqrt{ds^2}+\int_2^3 \sqrt{ds^2}=\Delta s_{12}+\Delta s_{23}.$$
Of course if your trajectory has constant velocity the integral simplifies and it looks like you're summing the two square roots.
A: For timelike intervals (the intervals that can be travelled by a particle), 
$d\tau^2 = g_{\alpha \beta} dx^{\alpha}dx^{\beta}$
Hence, $d\tau = \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}$
Thus, the total interval $\tau_{13} = \int_1^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}$
Therefore, 
$\tau_{13} = \int_1^2 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}} + \int_2^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}} = \tau_{12} + \tau_{23}$ . Always. 
The point is $\tau_{13}^2 \neq \tau_{12}^2+\tau_{23}^2$. 
Because, $d\tau^2 =  {g_{\alpha \beta} dx^{\alpha}dx^{\beta}} = g_{\alpha\beta}U^{\alpha}U^{\beta} d\lambda^2$
($\lambda$ is the affine parameter used to paramatrize the curve (path) along which the particle travels and $U^{\mu} := \dfrac{dx^{\mu}}{d\lambda}$)
Thus, in order to integrate, we first take the square root and then only perform the integration. There is no meaning to directly saying $\int_1^3 d\tau^2$. 
So, $\tau_{13}^2 = \big(\int_1^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}\big)^2$ NOT that $\tau_{13} = \sqrt{\int_1^3 {g_{\alpha \beta} dx^{\alpha}dx^{\beta}}}$. 
'Reduced Mathematics' Version:
Your intuition is right. $\tau_{13} = \tau_{12} + \tau_{23}$. Reason is simply that we first take square root of $d\tau^2$ and then integrate - not the other way around. Thus, $\tau_{13}^2 \neq \tau_{12}^2+\tau_{23}^2$ but  $\tau_{13} = \tau_{12} + \tau_{23}$.
