Differentiate the equation $\dot{\hat{x}}(t) = f(\hat{x}(t))$ with respect to $t$: \begin{align*} \ddot{\hat{x}}(t) = Jf_{\hat{x}(t)}\,\dot{\hat{x}}(t) = A(t)\,\dot{\hat{x}}(t). \end{align*} Thus $\dot{\hat{x}}(t)$ satisfies the variational equation. By the uniqueness of the fundamental matrix, $\dot{\hat{x}}(t) = F(t)\,\dot{\hat{x}}(0)$. Evaluating at $t = P$ and using periodicity $\dot{\hat{x}}(P) = \dot{\hat{x}}(0)$: \begin{align*} F(P)\,\dot{\hat{x}}(0) = \dot{\hat{x}}(P) = \dot{\hat{x}}(0). \end{align*} Since $\dot{\hat{x}}(0) \neq \mathbf{0}$ (the orbit has no fixed points), this shows $\dot{\hat{x}}(0)$ is an eigenvector of $F(P)$ with eigenvalue $1$.