Let $\pi:L(M)\to M$ by the projection onto the base. Define $f:M\to M$ by $f(x) := \pi(F(p))$, where $p\in \pi^{-1}(x)\subset L(M)$ is arbitrary. Since $F:L(M)\to L(M)$ is fibre-preserving (and so maps the fibre $\pi^{-1}(x)$ into some other fibre $\pi^{-1}(y)$), this definition is independent of the choice of $p\in \pi^{-1}(x)$, and so gives a function satisfying $f\circ \pi = \pi\circ F$.
Let $\pi:L(M)\to M$ by the projection onto the base. Define $f:M\to M$ by $f(x) := \pi(F(p))$, where $p\in \pi^{-1}(x)\subset L(M)$ is arbitrary. Since $F:L(M)\to L(M)$ is fibre-preserving (and so maps the fibre $\pi^{-1}(x)$ into some other fibre $\pi^{-1}(y)$), this definition is independent of the choice of $p\in \pi^{-1}(x)$, and so gives a function satisfying $f\circ \pi = \pi\circ F$.