Lemma 4.1.1 in D.J Saunders 's book (The geometry of jet bundles) says:
Let $(E,\pi,M)$ be a bundle, and let $p$ $\in M$. Suppose that $\phi$ and $\psi$ are section that satisfy $\phi(p)=\psi(p)$. Let $(x^i,u^\alpha)$ and $(y^i,v^\beta)$ be two adapted coordinate systems around $\phi(p)$, and suppose also that $$\frac{\partial( u^\alpha \circ \phi)}{\partial x^i}\bigg|_{p}=\frac{\partial( u^\alpha \circ \psi)}{\partial x^i}\bigg|_{p}$$ for $1 \leq i\leq m$ and $1 \leq \alpha \leq n$ then
$$\frac{\partial( v^\beta \circ \phi)}{\partial y^i}\bigg|_{p}=\frac{\partial( v^\beta \circ \psi)}{\partial y^i}\bigg|_{p}$$ for $1 \leq j\leq m$ and $1 \leq \beta \leq n$
To proof this he uses the relation
$$\frac{\partial( v^\beta \circ \phi)}{\partial y^j}\bigg|_{p}=\frac{\partial( v^\beta \circ \phi)}{\partial x^i}\bigg|_{p} \frac{\partial x^i }{\partial y^j}\bigg|_{p}=\left (\frac{\partial v^\beta }{\partial x^i}\bigg|_{\phi(p)}+\frac{\partial v^\beta }{\partial u^{\alpha}}\bigg|_{\phi(p)} \frac{\partial( u^\alpha \circ \phi)}{\partial x^i}\bigg|_{p} \right)\frac{\partial x^i }{\partial y^j}\bigg|_{p} \tag 1$$
What I am not seeing is why from $(1)$ we obtain $$\frac{\partial( v^\beta \circ \phi)}{\partial y^i}\bigg|_{p}=\frac{\partial( v^\beta \circ \psi)}{\partial y^i}\bigg|_{p}$$ for $1 \leq j\leq m$ and $1 \leq \beta \leq n$