How is $\partial/\partial t$ a vector? How is $\partial_t=\partial/\partial t$ a (Killing) vector in some coordinate system?
I know $\partial f/\partial t$ is the partial derivative of $f$ with respect to $t$.
But how about  $\partial/\partial t$? Is it still the partial derivative with respect to $t$? But of what?
 A: $\partial/\partial t$ is not a partial derivative. It is just notation. What you used to write as $\hat u_i$, or $\hat x_i$, is now written as
$$
\frac{\partial}{\partial x^i} \tag{1}
$$
In other words, $\partial/\partial x^i$ is a standard vector, but with a new, different notation. This notation is convenient for several reasons; for example, changing local coordinates looks like the chain rule (which ultimately is not really a coincidence, but the result of the fact that the tangent space is spanned by gradients).
Also, we usually define vectors as derivations, so that for a certain coordinate chart,
$$
v(f)\equiv\sum_i v_i \frac{\partial f}{\partial x^i} \tag{2}
$$
which is consistent with
$$
v=v_i\frac{\partial}{\partial x^i} \tag{3}
$$
where in $(2)$ the symbol $\partial$ denotes a true derivative, and in $(3)$ it denotes a basis vector.

If you will, in the old notation you can write the Killing vector as $k=(1,0,0,0)$ instead of $k=\partial/\partial t$. It is the exact same thing.
A: A vector $\xi^\mu$ is associated with the differential operator $\xi^\mu\partial_\mu$, which is $\partial_t$ for $\xi^\mu=\delta^\mu_0$ so  $\nabla^\mu\xi^\nu=0$. Thus $\nabla^\mu\xi^\nu+\nabla^\nu\xi^\mu=0$, making $\xi^\mu$ a Killing vector.
