I think that if you generalize that statement a little it becomes clearer (also the proof).
Let $G$ be any Lie group (not necessarily compact) with a closed subgroup $H$ and a metric (not necessarily positive definite) on $G$ which is $G$-left-invariant and $H$-right-invariant (not necessarily bi-invariant).
These conditions are equivalent to picking a metric (quadratic form) at $Lie(G)$ (the lie algebra of $G$, thought of as the tangent space at the identity) which is invariant under the Adjoint representation of $G$ restricted to $H$. You extend this metric from the identity to all of $G$ by left translations.
Example: $G=SL(2,R)$, $H=SO(2)$, with the Killing metric on $G$ (bi-invariant but not positive definite). In this case $G/H$ is the hyperbolic plane. Also any semi-simple $G$ with the Cartan-Killing metric and a maximal compact $H$ (then $G/H$ is called a symmetric space).
Another example is $G=SO(3)$, $H=SO(2)$ (standard embedding) with left-invariant metric which is not necessarily right-invariant, but $H$-right-invariant. This is a model for a rigid body motion whose ellipsoid of inertia is axially symmetric.
From these conditions you get that the metric descends to $G/H$ ($G$ modulo right traslations by $H$), and that left translations by $G$, which by definition act by isometries on $G$, descend to isometries on $G/H$ (since left and right translations commute, by associativity).
If you want the metric on $G/H$ to be riemannian (ie positive definite) then you need to ask that $Lie(G)/Lie(H)$ is positive definite. This holds in the examples above.
Next pick any vector $v\in Lie(G)$ and extend it to a right invariant vector field $X$ on $G$.
Exercise: the flow of $X$ is given by the action of the 1-parameter subgroup of $G$ generated by $v$, $g_t=exp(tv)$, acting by left translations on $G$.
Since left translations are isometries of $G$ it follows that $X$ is Killing. Since $X$ is right invariant it descends to a vector field $\tilde X$ on $G/H$ and the left translations by $g_t$ descend to the flow of $\tilde X$, which is by isometries, so $\tilde X$ is Killing.
Note that $v\in Lie(G)$ doesn't have to be transverse to $Lie(H)$. Picking $v\in Lie (H)$ generates Killing fields $\tilde X$ with fixed point $[H]\in G/H$.
Another comment is that this construction doesn't generate in general all the Killing fields on $G/H$.
Take for example $G$ compact with bi-invariant metric and $H$ trivial. The construction misses all the left-invariant vector fields on $G$ (generating right translations).