= Solution
A <Riemannian isometry> $F:(M,g)\to(N,h)$ is a <diffeomorphism> satisfying $F^*h=g$. A <local isometry> is a <local diffeomorphism> with the same pullback identity; equivalently $dF_p$ is an isometric <linear isomorphism> for every $p$. An <isometric immersion> into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.
Let $\phi_t$ be the <local flow> of the generating <vector field> $K$. The identity $\phi_t^*g=g$ differentiates to $\mathcal L_Kg=0$, where $\mathcal L$ is the <Lie derivative of a tensor field>. For arbitrary <vector fields> $X,Y$, <metric compatibility> and the <torsion-free connection> give
$$
\begin{aligned}
(\mathcal L_Kg)(X,Y)
&=K\langle X,Y\rangle-\langle[K,X],Y\rangle-\langle X,[K,Y]\rangle\\
&=\langle\nabla_XK,Y\rangle+\langle\nabla_YK,X\rangle.
\end{aligned}
$$
This proves the <Killing equation>. Conversely, a smooth <Killing field> has a <local flow>, and
$$
\frac{d}{dt}\phi_t^*g=\phi_t^*(\mathcal L_Kg)=0.
$$
Since $\phi_0$ is the identity, $\phi_t^*g=g$ wherever the flow is defined. Its inverse is $\phi_{-t}$ locally, so \b[the <local flow> consists of <local isometries>]. Neither direction assumes that $K$ is complete.
Put $H(X,Y)=\nabla_X(\nabla_YK)-\nabla_{\nabla_XY}K$ and $B(X,Y,Z)=\langle H(X,Y),Z\rangle$. These expressions are tensorial in $X,Y,Z$. Take the <covariant derivative> in direction $X$ of the <Killing equation>, subtracting the terms from differentiating $Y$ and $Z$. It follows that
$$
\boxed{B(X,Y,Z)+B(X,Z,Y)=0.}
$$
With the curvature convention $\mathcal R$ of Solution 1, the <torsion-free connection> also gives
$$
H(X,Y)-H(Y,X)=\mathcal R(X,Y)K.
$$
Define $C(X,Y,Z)=\langle\mathcal R(X,K)Y,Z\rangle$. It is skew in $Y,Z$ by skew-adjointness of the <Riemann curvature tensor>. The <first Bianchi identity> gives
$$
\mathcal R(X,K)Y-\mathcal R(Y,K)X=\mathcal R(X,Y)K,
$$
so $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:
$$
D(X,Y,Z)=D(Y,X,Z)=-D(Y,Z,X)=-D(Z,Y,X)
=D(Z,X,Y)=D(X,Z,Y)=-D(X,Y,Z).
$$
We have proved the <second covariant derivative of a Killing vector> identity, including its sign:
$$
\boxed{\nabla^2_{X,Y}K=\mathcal R(X,K)Y.}
$$
The printed identity uses the opposite <curvature-sign convention in Killing derivative identities>. If $R=-\mathcal R$, namely $R(X,Y)Z=\nabla_Y\nabla_XZ-\nabla_X\nabla_YZ+\nabla_{[X,Y]}Z$, the same result reads \b[$\nabla^2_{X,Y}K=R(K,X)Y$], exactly as printed. The convention must change with the sign; under $\mathcal R$, writing $\mathcal R(K,X)Y$ would be incorrect.
Finally let $H_0=K-\widetilde K$ have zero value and zero first <covariant derivative> at $p$. Along any <geodesic> issuing from $p$, the identity just proved says
$$
D_t^2H_0=\mathcal R(T,H_0)T=-\mathcal R(H_0,T)T.
$$
Thus $H_0$ is a <Jacobi field> with zero initial value and derivative. Uniqueness for this linear <ordinary differential equation> implies $H_0=0$ on each such <geodesic>, hence on a <convex normal neighborhood> of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is <connected>, therefore it is all of $M$. This proves \b[$K=\widetilde K$ everywhere] and the general statement that a <Killing field is determined by its value and first covariant derivative>.
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