= Solution
Treat $V$ as a derivation. For an arbitrary smooth function $f$ on $\mathcal N$, the definition of the <differential of a smooth map> and the <chain rule> give
$$
\begin{aligned}
(\phi_*V)[f]
&=V[f\circ\phi]
=V^i\partial_i(f(y(x)))\\
&=V^i\frac{\partial y^\alpha}{\partial x^i}\frac{\partial f}{\partial y^\alpha}.
\end{aligned}
$$
Comparison with $(\phi_*V)^\alpha\partial_\alpha f$ for all $f$ proves
$$
\boxed{(\phi_*V)^\alpha=\frac{\partial y^\alpha}{\partial x^i}V^i.}
$$
The Jacobian is an $n\times m$ matrix; \b[it need not be square or invertible].
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