What is time-like path in GR? My understanding: Given a metric, at each point of spacetime, there is a tangent vector u that maximize the quantity $g_{ab}u^au^b$, which is the proper-time length. Does it mean at each point, there is a specific velocity a clock (clockA) can have to be the quickest clicking clock? I know it doesn't make sense to say that some other observer at the same point with different velocity always see this clockA to click the quickest. 
I will appreciate it if you can point out my misunderstanding.
 A: Minkowski's 1906 paper introduced the space-time interval:
$$ \Delta s^{2} = -c^{2}\Delta t^{2} + \Delta x^{2} + \Delta y^{2} +\Delta z^{2} \tag{1} $$
or in differential fashion:
$$ ds^{2} = -c^{2}dt^{2} + dx^{2} + dy^{2} + dz^{2} \tag{2} $$
The well known Euclidean structure "stage" of a theoretical physical model had to change. The main (physical-some sort of philosofical) consequence of $(2)$ is that the metric $(2)$ (together, of course, Special Relativity axioms) "splits" the absolute sense of causality imposed by Euclidean metric (additionaly with postulates of newtonian framework); then you can quantify this notion inside three vector objects. Mathematicaly the "split" occurs because the expression of the metric is an scalar "produced" by an pseudo-norm. [considering metric signature $(-+++)$]:
\begin{cases}
ds^{2} =: \hspace{2mm}\mid \mid \mathbf{t} \mid \mid^{2} \hspace{2mm} \equiv \langle \mathbf{t},\mathbf{t} \rangle < 0  \hspace{10mm} time-like \hspace{1mm} vector\\
ds^{2} =: \hspace{2mm}\mid \mid \mathbf{l} \mid \mid^{2} \hspace{2mm} \equiv \langle \mathbf{l},\mathbf{l} \rangle = 0  \hspace{10mm} light-like \hspace{1mm} vector\\
ds^{2} =: \hspace{2mm}\mid \mid \mathbf{v} \mid \mid^{2} \hspace{2mm} \equiv \langle \mathbf{v},\mathbf{v} \rangle > 0  \hspace{10mm} space-like \hspace{1mm} vector\\
\end{cases}
or
\begin{cases}
t_{\mu}t^{\mu} = \eta_{\mu\nu}t^{\mu}t^{\nu} < 0  \hspace{10mm} t^{\mu} \hspace{1mm} is\hspace{1mm}a \hspace{1mm} time-like \hspace{1mm} vector\\
l_{\mu}l^{\mu} = \eta_{\mu\nu}l^{\mu}l^{\nu} = 0 \hspace{10mm} l^{\mu} \hspace{1mm} is\hspace{1mm}a \hspace{1mm} light-like \hspace{1mm} vector\\
v_{\mu}v^{\mu} = \eta_{\mu\nu}v^{\mu}v^{\nu} > 0 \hspace{10mm} v^{\mu} \hspace{1mm} is\hspace{1mm}a \hspace{1mm} space-like \hspace{1mm} vector\\
\end{cases}
Now, on the other hand, because the lorentzian structure (which receive physical meaning from the axioms of relativity, mainly due to the fact that $c$ is an (localy) absolute speed) we can say that the motion occurs in three ways, that one which massive particles perform; that one which light perform and then that one (unphysical) which a tachyon-like particle perfom. 
Respectively we can say that, massive particles perfom a path which in every point of the trajectory the four-velocity if a time-like vector; massless particles -such a photon- perfom a path which in every point of the trajectory the four-velocity if a light-like vector; tachyonic-like particles perfom a path which in every point of the trajectory the four-velocity if a space-like vector.
Furthermore, massive particle trajectories, (i.e. time-like paths) are parametrized by proper time. Light-like and space-like paths are parametrized by other affine parameters.
We could say that time-like paths obey the following geodesic equation:
$$\frac{\mathrm{du^{\mu}}}{\mathrm{d}\tau} \equiv \frac{\mathrm{d^{2}x^{\mu}}}{\mathrm{d}\tau ^{2}} = 0 \tag{3}$$
So, physically a time-like path is a massive particle's path. Mathematically is a path parametrized by proper time and the one who satisfies the condition $\eta_{\mu \nu}t^{\mu}t^{\nu} < 0$.
$$* * *$$
Now, General Relativity (GR) generalizes Special Relativity (SR) both in physical and mathmatical fashion. 
The metric is a generic first fundamental form defined by a $(0,2)-$tensor field introduced on the manifold $\mathcal{M}$:
$$(\mathcal{M}, \mathfrak{g})$$
$$ds^{2} = g_{\mu \nu}(x^{\gamma})dx^{\mu}dx^{\nu} =: \mathfrak{g}(\mathbf{k},\mathbf{k}) = g_{\mu \nu} \mathbf{d}x^{\mu} \otimes \mathbf{d}x^{\nu} $$

Remark: Note that a minkowski space (the space of special relativity) have, most of the time, the structure:
$$(\mathbb{R}^{4}, \mathfrak{\eta})$$
where $\eta$ is the Minkowski metric tensor field.
also, look this question (first answer) for more intuition on metric tensors: 
Is a metric tensor field the same thing as $ds² = -dt² + dx²+ dy² + dz²$? 

Now, the "split" imposed by lorentzian geometry is still valid in GR. But now, the metric is a generic one and because of that, the equation of motion which massive particles satisfy are more general:
$$\frac{\mathrm{d^{2}x^{\mu}}}{\mathrm{d}\tau ^{2}} + \displaystyle \Gamma^{\mu}\hspace{0.5mm}_{\nu \gamma}\frac{\mathrm{dx^{\nu}}}{\mathrm{d}\tau}\frac{\mathrm{dx^{\gamma}}}{\mathrm{d}\tau} = 0 \tag{4}$$
You mentioned extremals and proper length. So, both $(3)$ and $(4)$ are equations derived from a particular integral, the only difference is in the metric tensor used (and of course in causal character of the vectors).
A time-like path of a particle, between two events $A$ and $B$, is (a geodesic!) the one given by:
$$\delta \int_{A}^{B} ds = 0 \tag{5}$$
Then fo SR,
$$\delta \int_{A}^{B} ds = 0 \iff \delta \int_{A}^{B} \sqrt{\eta_{\mu \nu} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau}} = 0 \iff \delta \int_{A}^{B} d\tau \sqrt{\eta_{\mu \nu} dx^{\mu} dx^{\nu}} = 0 \implies \tag{7} $$
$$\implies \frac{\mathrm{d^{2}x^{\mu}}}{\mathrm{d}\tau ^{2}} = 0 $$
And for GR:
$$\delta \int_{A}^{B} ds = 0 \iff \delta \int_{A}^{B} \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau}} = 0 \iff \delta \int_{A}^{B} d\tau \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}} = 0 \implies \tag{7} $$
$$\implies \frac{\mathrm{d^{2}x^{\mu}}}{\mathrm{d}\tau ^{2}} + \displaystyle \Gamma^{\mu}\hspace{0.5mm}_{\nu \gamma}\frac{\mathrm{dx^{\nu}}}{\mathrm{d}\tau}\frac{\mathrm{dx^{\gamma}}}{\mathrm{d}\tau} = 0 \tag{8} $$
And in both $(7)$ and $(8)$, the tangent components $\frac{dx^{\mu}}{d\tau}$ are components of a time-like vector (four-velocity). 
Then in GR a quite good way to define a time-like path, is the trajectory given by $(4)$.
