The question is literally this simple. I would like to know, in not extremely complicated terms, in what sense does a Killing vector field generate an isometry. I am somewhat familiar with the concept of Lie derivatives.
The explanations I came across seem not so clear. Maybe I don't have enough background for those extremely formal explanations. I would appreciate any help!
On
In general, a vector field $X$ on a manifold $M$ generates (using Picard-Lindelöf theorem for ordinary differential equations) a local flow $(\phi_t\colon M\to M)_t$ ($t$ being small enough) by: $$\frac{\operatorname{d}\phi_t(x)}{\operatorname{d}t}_{\vert s}=X(\phi_s(x)),\phi_0(x)=x.$$ Geometrically, $\phi_t(x)$ is a obtained following $X$ from $x$ during a time $t$.
When $(M,g)$ is a Riemannian manifold and $X$ is a Killing vector field, $\phi_t$ is an isometry.
The reverse is in some ways a bit more clear; given a 1-parameter family of isometries $\phi_t$, their infinitesimal generator $$ X_p := \frac{d}{dt}\phi_t(p)|_{t = 0} $$ is a Killing field which generates $\phi_t$. The reason $X$ is Killing because $L_X g = 0$, since by assumption $\phi_t$ is an isometry for each $t$ (hence $\phi_t^* g = g$).
Conversely, let $X$ be a smooth vector field on $M$. Let $\phi_t$ be its flow, so $\frac{d}{dt}\phi_t(p) = X(\phi_t(p))$ and $\phi_0(p) = p$. Then this flow (which $X$ "generates") is an isometry iff $X$ is Killing. We have discussed the forward direction above. So let us assume $X$ is Killing. This means that the Lie derivative of the metric is everywhere zero: $L_X g \equiv 0$. Hence the "temporal" function $(\phi_t^*g)_p$ is constant in time at every point $p$ on the manifold. In particular $\phi_t^* g = g$, i.e. $\phi_t$ is an isometry.
For more on this last fact, you could check out Lee's Smooth Manifolds book. It essentially boils down to showing that $$ \frac{d}{dt}((\phi_t^* g)_p)|_{t = s} = (\phi_s^*(L_X g))_p. $$