What's the point of this killing vector notation? Reading Sean Carroll's spacetime and geometry he says

If $x^{\sigma_*}$ is the coordinate which ${\mu\nu}$ is independent of, let us consider the vector $\partial_{\sigma_*}$ which we label as 
  $$K=\partial_{\sigma_*}\, , \tag{3.169}$$ 
  which is equivalent in component notation to
  $$K^\mu=(\partial_{\sigma_*})^\mu = \delta^\mu_{\sigma_*}. \tag{3.170} $$

Now I ask you, why call a vector that for example looks like this $(1,0,0,0)$ as $\partial_0$?
Is there a point and a reason behind this notation which I always found pretty confusing? 
For example before reading Sean Carroll the times I read a phrase like the Killing vector is $\partial/\partial t$ I thought it was something like $(\partial_t,0,0,0)$ and beside that I was never sure if it was really $(\partial_t,0,0,0)$ or if it was a spacelike vector with a component equal to $\partial_t$
 A: Just think about ordinary vectors. The vector $\mathbf{r} = \hat{\mathbf{x}}$ has components $(1, 0, 0)$ in Cartesian coordinates. This is because the components are the numbers you have to multiply the unit vectors with to recover the vector you want,
$$\mathbf{r} = 1 \hat{\mathbf{x}} + 0 \hat{\mathbf{y}} + 0 \hat{\mathbf{z}}.$$
The exact same thing is true here, except that the unit vectors are $\partial_t$, $\partial_x$, $\partial_y$, and $\partial_z$. Again, the components of vectors are numbers, not vectors themselves.
A: It comes from the formal definition of a vector on a manifold. Given a manifold $\mathcal{M}$ and a chart $(U,x^{\mu}:U\to\mathbb{R}^n)$ one defines a curve $c$ and a function $f$ both differentiable
$$
c:\mathbb{R}\to\mathcal{M}\,,\qquad f: \mathcal{M}\to\mathbb{R}\,.
$$
Since $f\circ c$ is a function $\mathbb{R}\to\mathbb{R}$ one can define its first derivative
$$
\left.\frac{\mathrm{d}f(c(t))}{\mathrm{d}t}\right|_{t=0}  = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t} \frac{\partial f}{\partial x^\mu}\equiv V f\,.
$$
The differential operator $V = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t} \frac{\partial }{\partial x^\mu}$ is what we call a vector.
The nice thing is that such an expression is coordinate independent: $Vf$ only depends on the choice of the curve $c$ (modulo an obvious equivalence relation) and $f$. On the other hand, when we strip the $\partial/\partial x^\mu$ and define
$$
V^\mu = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t}\,,
$$
we are introducing the dependence on the chosen coordinate frame $x^\mu$.
A: There are several different ways to define tangent vectors, eg in terms of coordinates and transformation laws, or as equivalence classes of differentiable curves in first-order contact. In differential geometry, we tend to identify tangent vectors with their corresponding directional derivatives, which can be used as a third way to introduce the tangent space.
Writing the basis vectors induced by a coordinate chart as partial derivatives makes sense under that identification.
For example, let $\Psi = \Psi(x^1,\dots,x^n)$ be a parametrization of our manifold $M$. Then, each of the $n$ coordinates will induce a tangent vector $\partial/\partial x^i$, corresponding to the directional derivative
$$
\frac{\partial}{\partial x^i}f = \frac{d}{d\tau}\Big|_{\tau=0} (f\circ\Psi)(x^1,\dots,x^{i-1},x^i + \tau,x^{i+1},\dots,x^n)
$$
of a function $f:M\to\mathbb R$.
In terms of this basis, an arbitrary vector $X$ can then be expanded as
$$
X = \sum_{i=1}^n X^i \frac{\partial}{\partial x^i}
$$
For $x^0 = t$ and $X=\partial/\partial t$, we would then have
$$
X = 1\cdot \frac{\partial}{\partial t} + 0\cdot\frac{\partial}{\partial x^1} + 0\cdot\frac{\partial}{\partial x^2} + 0\cdot\frac{\partial}{\partial x^3}
$$
ie
$$
(X^\mu) = (1,0,0,0)
$$
but not
$$
(X^\mu) = (\partial/\partial t,0,0,0)
$$
