Calculate the pushforward of a vector field Consider the vector fields
$X=xy\frac{\partial}{\partial x}+x^{2}\frac{\partial}{\partial z}$, and $Y=y\frac{\partial}{\partial y}$ on $R^{3}\rightarrow R$, and $f(x,y,z)=x^{2}y$. 
Compute:  $f_{*}(X_{(1,1,0)})$
There is a similar question posted here, I read it but it did not answer my question, I also searched for the formula but I don't think it helped..
I got a solution saying:
$f_{*}(X_{(1,1,0)})=(\frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \frac{\partial f}{\partial z})_{(1,1,0)}\begin{pmatrix} 
1\\
0\\
1\end{pmatrix}=2\frac{d}{dt} \mid_{1}$, where $t$ denotes the canonical coordinates on $R$.
I understand that $(\frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \frac{\partial f}{\partial z})_{(1,1,0)}$ is the Jacobian matrix of $f$ at point $(1,1,0)$, and I found that if I plug the point $(1,1,0)$ into $X$, and it will give me $\frac{\partial}{\partial x}+0 \frac{\partial}{\partial y}+\frac{\partial}{\partial z}$, so it will give me $\begin{pmatrix} 
1\\
0\\
1\end{pmatrix}$. and thus multiplying two matrices will get me $2$
But why all of those become together? and why it produced 
$2\frac{d}{dt} \mid_{1}$, and what is the canonical coordinates on $R$?
 A: If you have a differentiable map $f:M\to N$, then its differential $Df$ at a given point $p$ is a linear map $T_pM\to T_{f(p)}N$. So far, so good.
Now the linear map $Df$ is represented by the Jacobian matrix given a system of coordinates. This matrix takes as input a $3$-coordinate vector and outputs a $1$-coordinate vector, which is a little bit more than just a real number because it lives in the tangent space to $\mathbb{R}$ at a particular point. Also, note that for a matrix to make sense you need to choose coordinates in the domain, and in the codomain. So there you are, $\frac{\partial}{\partial t}|1$ is the basis vector of the tangent space to $\mathbb{R}$ at the point $1$.
If your vector in $T_pM$ is given by $v=(v_x,v_y,v_z)$ then $Df_x(v)$ is by definition the dot product $\left(\frac{\partial f}{\partial x}|_p\,\,\frac{\partial f}{\partial y}|_p\,\,\frac{\partial f}{\partial z}|_p\right)\cdot(v_x\,\,v_y\,\,v_z)$ multiplied by this basis vector.
The canonical coordinates in $\mathbb{R}$ is simply the coordinate system induced by considering the identity map as a coordinate map.
