So let me state the lemma first
Lemma 40:If $f,h:(M,g)\rightarrow \mathbb{R}$ are $C^2$ functions such that $f(p)=h(p)$ and $f(x)\geq h(x)$ for all $x$ near $p$, then
$$ \begin{align} \nabla f(p)&=\nabla h(p)\\ \text{Hess}f\mid_p&\geq\text{Hess}h\mid_p\\ \Delta f(p)&\geq \Delta h(p) \end{align} $$
So we know that $df(\dot{\gamma}(0))=dh(\dot{\gamma}(0))$ for all curves $\gamma :(-\delta,\delta)\rightarrow M$ with $\gamma (0)=p$ by simple calculus. Then he states that this implies
$$\text{Hess}f(\dot{\gamma}(0),\dot{\gamma}(0))\geq\text{Hess}h(\dot{\gamma}(0),\dot{\gamma}(0))$$
How does he get this conclusion. Could someone please explain it to me. It seems that since $f(x)\geq h(x)$ near $p$ we have $$\dot{\gamma}(0)\dot{\gamma}(0)f(p)\geq\dot{\gamma}(0)\dot{\gamma}(0)h(p)$$ but I'm unsure if this is what Peterson is doing. If someone could explain Peterson's proof I would appreciate it.
Note that since $\gamma$ is a geodesic, we have
$$ {\rm Hess} f (\dot \gamma, \dot \gamma) := \dot \gamma (\dot \gamma f)-(\nabla _{\dot \gamma} \dot \gamma)f=\dot \gamma (\dot \gamma f).$$
Thus it suffices to see that $\dot \gamma (\dot \gamma f)|_p\ge\dot \gamma (\dot \gamma h)|_p$, which is true because $\dot \gamma (\dot \gamma f)|_p = (f \circ \gamma)''(0)$, and (as Petersen puts it) "simple calculus" tells us that $(f \circ \gamma)''(0) \ge (h \circ \gamma)''(0)$.