As we know ,if $\varphi:M\rightarrow N$ is a differentiable map, then ,the map $\varphi$ will induces a linear map $\varphi_*:T_xM\rightarrow T_{\varphi(x)}N$ ,up to $(\varphi_*X)(f)=X(f\circ\varphi)$,$X\in T_xM$.

but, how to represent the $\varphi_*X$ in local coordinate? I mean that :

$\{x_i\}$ and $\{y_j\}$ are local coordinates of some neighborhoods of $x$ and $\varphi(x)$,$X=X^i\frac{\partial }{\partial x^i}$, then how to represent $\varphi_*X$ by the local coordinates?


It's just the jacobian matrix $(\frac{\partial \phi^k}{\partial x^i})$, where $\phi^k=y^k\circ\phi$. The push forward/differential generalizes the derivative from real analysis.

To see why, note that since $\phi_*$ is linear, it is completely determined by its action on the basis vectors. If $\{\frac{\partial}{\partial x^i}_p\}$ and $\{\frac{\partial}{\partial y^j}_{\phi(p)}\}$ are the respective chart induced basis for $T_pM$ and $T_{\phi(p)}N$, then to determine $\phi_*$, we just need to know $\phi_*(\frac{\partial}{\partial x^i}_p)$. Since $\phi_*(\frac{\partial}{\partial x^i}_p)\in T_{\phi(p)}N$, we have $\phi_*(\frac{\partial}{\partial x^i}_p)=\sum_jb_i^j\frac{\partial}{\partial y^j}_{\phi(p)}$. Seeing how both sides act on the function $y^k$, we find $b_i^k=(\frac{\partial y^k\circ\phi}{\partial x^i}_p)$


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