= Solution
For a scalar, the <Lie derivative of a function> is its <directional derivative>:
$$
\mathcal L_\xi f=\xi^a\partial_af.
$$
The <Lie derivative of a vector field> is their <Lie bracket of vector fields>:
$$
(\mathcal L_\xi X)^a=\xi^b\partial_bX^a-X^b\partial_b\xi^a.
$$
For a <torsion-free connection>,
$$
(\nabla_\xi X-\nabla_X\xi)^a=\xi^b\partial_bX^a-X^b\partial_b\xi^a+\Gamma^a{}_{bc}(\xi^bX^c-X^b\xi^c).
$$
The last term vanishes because the <connection coefficients> are symmetric in $b,c$. This proves \b[$\mathcal L_\xi X=\nabla_\xi X-\nabla_X\xi$].
For the <metric tensor> statements use the <Levi-Civita connection>, which also has <metric compatibility>. The <tensor derivation> property of the <Lie derivative of a tensor field> gives
$$
(\mathcal L_\xi g)(X,Y)=\xi[g(X,Y)]-g(\mathcal L_\xi X,Y)-g(X,\mathcal L_\xi Y).
$$
Expanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains is
$$
(\mathcal L_\xi g)(X,Y)=g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=2\xi_{(a;b)}X^aY^b.
$$
Since $X,Y$ are arbitrary, the <Killing equation> is
$$
\boxed{\mathcal L_\xi g=0\quad\Longleftrightarrow\quad\xi_{(a;b)}=0.}
$$
<Metric compatibility> is needed here, even though symmetry alone sufficed for the vector identity. In general the <Lie derivative of a metric with nonmetricity> contains a term $\xi^c\nabla_cg_{ab}$. For example, with the <metric tensor> $g=e^{2x}dx^2$, zero <connection coefficients> and $\xi=\partial_x$, the <Lie derivative of a tensor field> is $2e^{2x}$, whereas $2\nabla_x\xi_x=4e^{2x}$. The usual <metric connection> interpretation of the question supplies the required compatibility.
For the <Riemann curvature tensor> calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}R^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}R_{abce}$, it means
$$
[\nabla_c,\nabla_d]X^a=-R^a{}_{bcd}X^b.
$$
Thus the <commutator> on a <covector> has the opposite sign. Write $K_{abc}=\xi_{a;bc}=\nabla_c\nabla_b\xi_a$. Taking a <covariant derivative> of the <Killing equation> makes $K$ antisymmetric in its first two indices, while the <Ricci identity> gives
$$
K_{abc}-K_{acb}=R_{adbc}\xi^d.
$$
Combining this with the first-two-index antisymmetry gives the three relations
$$
K_{abc}+K_{cab}=R_{adbc}\xi^d,\quad K_{cab}+K_{bca}=R_{cdab}\xi^d,\quad K_{bca}+K_{abc}=R_{bdca}\xi^d.
$$
Add the first and third and subtract the second. Pair symmetry and the <first Bianchi identity> imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. Consequently
$$
K_{abc}=-R_{abcd}\xi^d,\qquad\boxed{\xi_{b;ca}=-R_{bca}{}^d\xi_d.}
$$
This is the <second covariant derivative of a Killing vector> in the paper's convention; its sign is tied to the declared <curvature-sign convention in Killing derivative identities>, not to raising a particular index.
Set $L_{ab}=\xi_{a;b}$. Along any <smooth curve> with <tangent vector> $V$, the <chain rule> and the just-proved identity give the <Killing transport equations>
$$
\boxed{\nabla_V\xi_a=L_{ab}V^b,\qquad\nabla_VL_{ab}=-R_{abc}{}^dV^c\xi_d.}
$$
Choose a <parallel-propagated orthonormal frame> along the curve. These become a homogeneous <linear system of ordinary differential equations> for the components of $(\xi,L)$. Uniqueness of its <initial value problem> implies that zero initial data remain zero. Any point in the same <connected component> as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore \b[zero value and zero first derivative at $P$ force the <Killing field> and its derivative to vanish throughout that component], with no completeness assumption. On a <manifold> with one <connected component> this is the requested global result. <Connectedness> cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.
On a <pseudo-Riemannian manifold> of dimension $n$ with one <connected component>, the map $\xi\mapsto(\xi_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The <dimension bound for Killing vector fields> is
$$
\boxed{\dim\mathfrak{kill}(M,g)\leq\frac{n(n+1)}2.}
$$
It is attained in flat space by $\xi_a=A_a+B_{ab}x^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or <Lorentz transformations>. The four-dimensional maximum is \b[10]. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.
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