Expression in D.J Saunders 's book (The geometry of jet bundles) 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 
$$\frac{\partial( 
 v^\beta \circ \phi)}{\partial y^j}\bigg|_{p}=\frac{\partial( 
 v^\beta \circ \phi)}{\partial x^i}\bigg|_{p} \frac{\partial 
 x^i }{\partial y^j}\bigg|_{p}=\left (\frac{\partial 
 v^\beta }{\partial x^i}\bigg|_{\phi(p)}+\frac{\partial 
 v^\beta }{\partial u^{\alpha}}\bigg|_{\phi(p)} \frac{\partial( 
 u^\alpha \circ \phi)}{\partial x^i}\bigg|_{p} \right)\frac{\partial 
 x^i }{\partial y^j}\bigg|_{p} $$
My problem  here is why $$\frac{\partial( 
 v^\beta \circ \phi)}{\partial x^i}\bigg|_{p} =\left (\frac{\partial 
 v^\beta }{\partial x^i}\bigg|_{\phi(p)}+\frac{\partial 
 v^\beta }{\partial u^{\alpha}}\bigg|_{\phi(p)} \frac{\partial( 
 u^\alpha \circ \phi)}{\partial x^i}\bigg|_{p} \right)$$
Shouldn't it be $$ \frac{\partial( 
 v^\beta \circ \phi)}{\partial x^i}\bigg|_{p}=\frac{\partial( 
 v^\beta \circ u^{-1} \circ u\circ \phi)}{\partial x^i}\bigg|_{p}=\frac{\partial 
 v^\beta }{\partial u^{\alpha}}\bigg|_{\phi(p)} \frac{\partial( 
 u^\alpha \circ \phi)}{\partial x^i}\bigg|_{p} $$
?
So why the extra factor $$\frac{\partial 
 v^\beta }{\partial x^i}\bigg|_{\phi(p)}$$
appears?
 A: Since $(x^k,u^\alpha)$ are local coordinates on $E$ around $\phi(p)$, the chain rule is $$\frac{\partial(v^\beta \circ \phi)}{\partial x^i}\bigg\vert_p = \sum_{k=1}^m \bigg(\frac{\partial v^\beta}{\partial x^k}\bigg\vert_{\phi(p)}\cdot\frac{\partial(x^k \circ \phi)}{\partial x^i}\bigg\vert_p\bigg) + \sum_{\alpha=1}^n \bigg(\frac{\partial v^\beta}{\partial u^\alpha}\bigg\vert_{\phi(p)}\cdot\frac{\partial(u^\alpha \circ \phi)}{\partial x^i}\bigg\vert_p\bigg).$$
The first sum simplifies because $x^k \circ \phi = x^k$ (slight abuse of notation: the first $x^k$ is the function on $E$ and the second is the function on $M$).

It is just the multivariable chain rule. If you prefer, you can write in these coordinates $$\phi(x_1,\ldots,x_m) = (x_1,\ldots,x_m,\phi^1(x_1,\ldots,x_m),\ldots,\phi^n(x_1,\ldots,x_m)),$$ where $\phi^\alpha = u^\alpha \circ \phi$; then the formula is $$\frac{\partial(v^\beta \circ \phi)}{\partial x^i}\bigg\vert_p = \sum_{k=1}^m \bigg(\frac{\partial v^\beta}{\partial x^k}\bigg\vert_{\phi(p)}\cdot\frac{\partial x^k}{\partial x^i}\bigg\vert_p\bigg) + \sum_{\alpha=1}^n \bigg(\frac{\partial v^\beta}{\partial u^\alpha}\bigg\vert_{\phi(p)}\cdot\frac{\partial \phi^\alpha}{\partial x^i}\bigg\vert_p\bigg).$$
