Parallel transport is a linear isomorphism

For real $c,d$, the linear combination $c{V}_v+d{V}_w$ solves the homogeneous equation and has initial value $cv+dw$. Uniqueness identifies it with $V_{cv+dw}$. Evaluating at $t$ proves linearity of $P_{\gamma ;s,t}$. Restarting the same section at $t$ and using uniqueness gives $P_{\gamma ;t,s}{P}_{\gamma ;s,t}v=v$ for every $v$; interchanging $s,t$ gives the other inverse identity.

In a frame, put $h(u)=a+b-u$. If $v^{\mathrm{\prime }}+\omega (\dot{\gamma })v=0$, then $(v\circ h{)}^{\mathrm{\prime }}+\omega ((\gamma \circ h{)}^{\mathrm{\prime }})(v\circ h)=h^{\mathrm{\prime }}(v^{\mathrm{\prime }}+\omega (\dot{\gamma })v)\circ h=0$. Thus reversing a parallel section is parallel along the reversed curve, continuously at its corners. Its endpoint values are interchanged, proving $P_{\bar{\gamma }}=P_{\gamma ;b,a}=P_{\gamma }^{-1}$. For $a=b$ these are identity maps; for rank zero they are the unique isomorphism of zero spaces. Zero vectors and rank one are already included in the linear calculation.