My question refers to the proof of the following theorem, but it may suffice to just skip the theorem and continue with the problematic taylor expansion $(\ast)$:
Let $f(t,y)$ and the single step method $\varphi(t,u,h)$ be sufficiently often continuously differentiable and let the local error have an expansion of the form $$ le(t,h) = d_{p+1}(t)h^{p+1} + d_{p+2}(t)h^{p+2} + \cdots + d_{N+1}(t)h^{N+1} + \mathcal{O}(h^{N+2}). $$ Then, the global error $e_h(t)$ after $n$ steps with size $h$ at $t^* = t_0 + nh$ has an asymptotic expansion $$e_h(t^*) = e_p(t^*)h^p + e_{p+1}(t^*)h^{p+1} + \cdots + e_N(t^*)h^N + E_{N+1}(t^*,h)h^{N+1},$$ where $E_{N+1}(t^*,h)$ for $0<h\leq h_0$ is bounded.
In the proof, a Taylor expansion is carried out
$$\varphi(t,y(t),h) - \varphi(t,y(t) - e_p(t)h^p,h) = \frac{\partial}{\partial y} \varphi(t,y(t),h)e_{p}h^p + \mathcal{O}(h^{2p}) \tag{$\ast$} $$
which I don't understand. Apparently, this is a Taylor expansion of a function of $h$ at $0$. But if I consider the function
$\varphi \circ g$, where $$g(h) := (t,y(t) - e_p(t)h^p,h)^T$$ then the Jacobian is $$J_g(h) = (0,-pe_p(t)h^{p-1},1)^T$$ wheras the Jacobian of $\varphi$ is $$J_{\varphi}(x,y,h) = (\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{ \partial h})$$ The chainrule would yield $J_{\varphi \circ g} = J_\varphi \cdot J_g$ which does not correspond to the result at $(\ast)$.
Do you have any suggestions on how to interpret $(\ast)$ correctly?
The Taylor expansion is for $$ φ(t,y(t),h)−φ(t,y(t)−s,h)=\frac{∂}{∂y}φ(t,y(t),h)s+O(s^2) $$ where $s=e_p(t)h^p$ so that $O(s^2)=O(h^{2p})$.