How to prove this formula involving the push-forward funtion Let $f:M^{n} \to N^{m}$, and suppose that $(x,U)$ and $(y,V)$ are coordinate systems around $p$ and $f(p)$, respectively, then I want to prove that 
$$f_{*} (\frac{\partial}{\partial x_i} \big{|}_p )=\sum_{j=1}^{m} \frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p) · \frac{\partial}{\partial y^{j}} \big{|}_{f(p)} $$
So then I think I can get a formula just extending by linearity for
$$f_{*} (\sum_{i=1}^{n} a^{i} \frac{\partial}{\partial x_i} \big{|}_p )$$
in terms of $\frac{\partial}{\partial y^{j}} \big{|}_{p}$ (I don't know if what I am thinking of this is right or this fact involves another proof). So then I am trying to use that
$$\frac{\partial(g \circ f)}{\partial x^{i}}(p)=\sum_{j=1}^{m}\frac{\partial(g )}{\partial y^{j}}(f(p))·\frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p)$$ 
because I have a similar expression but I can't figure how can I compute $f_{*} (\frac{\partial}{\partial x_i} \big{|}_p )$.
Can someone provide a detailed proof of this result please?

Attempt

We know that $f_{*}v(·)=v(· \circ f)$ then in this particular case we have $v=\frac{\partial}{\partial x_i} \big{|}_p $ so then using that
$$\frac{\partial(g \circ f)}{\partial x^{i}}(p)=\sum_{j=1}^{m}\frac{\partial(g )}{\partial y^{j}}(f(p))·\frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p)$$ 
we get that 
$$\frac{\partial}{\partial x_i} \big{|}_p (· \circ f)=\sum_{j=1}^{m} \frac{\partial(y^{j} \circ f)}{\partial x^{i}}(p) · \frac{\partial}{\partial y^{j}} \big{|}_{f(p)}$$
Am I right? and for the generalization what can be done?
Thanks in advance.
 A: If $\phi_M,\ \phi_N$ are charts then we have $G :=\phi_N\circ f \circ \phi_M^{-1} : {\bf R}^n\rightarrow {\bf R}^m$ so that $G_\ast \partial_{x_i}= \frac{\partial G^j}{\partial x_i} \partial_{y_j} $ If $E_i:=d\phi_M^{-1} \partial_{x_i}$ and $F_j:=d\phi_N^{-1}\partial_{y_j}$, then $$ f_\ast E_i= \frac{\partial G^j}{\partial x_i}  F_j $$
For $f_\ast \sum_i a_i\partial_{x_i}$, we need the argument about tangent space of manifold : If $d\phi_M^{-1},\ d\phi_N$ is linear, then since $dG$ is linear, we are done 
If $c$ is curve in ${\bf R}^n$, then $v:=d\phi_M^{-1}\ c'(0)\in
T_aM$ where $a=\phi_M^{-1}\circ c(0)$ If $h$ is a function on $M$,
then $$vh := \frac{d}{dt} h\circ \phi_M^{-1}\circ c $$
That is tangent vector $v$ is viewed as a differential operator. And
 $$ w:= \frac{d}{dt} \phi_M^{-1}\circ c_2 \Rightarrow
 Cv+w = \frac{d}{dt} \phi_M^{-1} ( c(Ct) + c_2(t)-c_2(0)  ) $$
$$ Cv+w = d\phi_M^{-1} \ (Cc' + c_2') $$
And $$ (Cv+w)h= d ( h\circ \phi_M^{-1} )\ ( Cc' + c_2')
=  Cd ( h\circ \phi_M^{-1} )\ c'+ d ( h\circ \phi_M^{-1} )\ c_2'
$$
$$ = C vh + w h $$
That is $d\phi_M^{-1} \ (Cc' + c_2') =Cd\phi_M^{-1} \ c'+
d\phi_M^{-1} \ c_2' $
