Space
To describe the direction of a path through space, you use a forward unit tangent vector: a unit vector which is tangent to the path, and points forward rather than backward.
If you have a path $\mathbf{x}^c$, parameterized by some variable $s$ that increases as you go forward, you can find the unit tangent vector $\mathbf{u}^c$ at each point using the formula
$$\mathbf{u}^c = \left. \frac{d\mathbf{x}^c}{ds} \right/ \sqrt{g_{ab} \frac{d\mathbf{x}^a}{ds} \frac{d\mathbf{x}^b}{ds}}.$$
Here, $g_{ab}$ is the metric of space. By calculating $g_{cd} \mathbf{u}^c \mathbf{u}^d$ from the formula, you can confirm that $\mathbf{u}^c$ is a unit vector.
You can simplify this formula by using an arc length parameterization. That means choosing a path parameter $\sigma$ for which
$$g_{ab} \frac{d\mathbf{x}^a}{d\sigma} \frac{d\mathbf{x}^b}{d\sigma} = 1.$$
In terms of an arc length parameter $\sigma$, the forward unit tangent vector has the elegant expression
$$\mathbf{u}^c = \frac{d\mathbf{x}^c}{d\sigma}.$$
Spacetime
To describe the direction of a timelike path through spacetime, you use a forward unit tangent vector (image source): a unit vector which is tangent to the path, and points forward rather than backward. (We typically orient timelike paths so that going forward along the path also means going forward in time.)
If you have a timelike path $\mathbf{x}^c$, parameterized by some variable $t$ that increases as you go forward, you can find the unit tangent vector $\mathbf{u}^c$ at each point using the formula
$$\mathbf{u}^c = \left. \frac{d\mathbf{x}^c}{dt} \right/ \sqrt{g_{ab} \frac{d\mathbf{x}^a}{dt} \frac{d\mathbf{x}^b}{dt}}.$$
Here, $g_{ab}$ is the metric of spacetime. Since the path is timelike, the number under the square root is always positive.
You can simplify this formula by using a proper time parameterization. That means choosing a path parameter $\tau$ for which
$$g_{ab} \frac{d\mathbf{x}^a}{d\tau} \frac{d\mathbf{x}^b}{d\tau} = 1.$$
In terms of a proper time parameter $\tau$, the forward unit tangent vector has the elegant expression
$$\mathbf{u}^c = \frac{d\mathbf{x}^c}{d\tau}.$$
World-velocity and spatial velocity
As you may have guessed by now, the world-velocity is another name for the forward unit tangent vector of a worldline. To see how the world-velocity is related to the familiar spatial velocity from Galilean mechanics, suppose spacetime is flat, so we can choose coordinates $t$, $x$, $y$, $z$ with
$$g_{ab} = dt_a\,dt_b - \left(dx_a\,dx_b + dy_a\,dy_b + dy_a\,dz_b\right).$$
Writing the world-velocity in terms of these coordinates as
$$\mathbf{u}^c = \left[\begin{array}{c}u_t \\ u_x \\ u_y \\ u_z\end{array}\right],$$
we can express the spatial velocity as a slope:
$$\frac{1}{u_t} \left[\begin{array}{c} u_x \\ u_y \\ u_z \end{array}\right].$$
To get a more explicit expression for the world-velocity, let's use the time coordinate $t$ as our path parameter, giving
$$\mathbf{u}^c = \frac{1}{\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}}\left[\begin{array}{c} 1 \\ dx/dt \\ dy/dt \\ dz/dt \end{array}\right].$$
When we take the slope, that square-root normalization factor cancels out, and we see that the spatial velocity is
$$\frac{1}{1} \left[\begin{array}{c} dx/dt \\ dy/dt \\ dz/dt \end{array}\right],$$
just like we'd expect.