So, for a scalar field $T(x,y,z)$, the derivative along $d\vec l$ is given by $$\frac {dT}{|d\vec l|} = |\vec \nabla T| \cos\theta$$where $\theta$ is the angle between $\vec \nabla T$ and $d\vec l$
For a vector field $\vec V (x,y,z)$, I understand that $\vec \nabla . \vec V$ and $\vec \nabla \times \vec V$ give the Divergence and the Curl respectively.
But, is there a way in which $\vec \nabla$ can act on $\vec V$ to give an expression for $\frac {d \vec V}{|d\vec l|}$, the directional derivative of $\vec V$ along $d\vec l$?
PS: I've only just started to learn vector calculus, so pardon me if this question comes out as silly.
The nabla operator is not the correct tool for vector fields. In this case, it is better to recognize that the gradient $\nabla T$ is just a special case of the total differential $\mathrm DT$, which is just the Jacobian of $T$. The directional derivative in $x_0$ in direction $v$ is then
$$\mathrm DT(x_0)\cdot e_v,$$
where $e_v$ is the unit vector in $v$-direction. This will be a vector, since the Jacobian $\mathrm DT$ is a matrix, but this is expected, since $T$ is vector valued. Going back to the special case of a scalar field, where $\mathrm DT=\nabla T$, this becomes
$$\mathrm DT\cdot e_v=\nabla T\cdot e_v=\vert\nabla T\vert\vert e_v\vert\cos\theta=\vert\nabla T\vert\cos\theta,$$
which is exactly what you started with.