Show that the coordinates of $X = (d\phi)(\tilde{X})$ are $ X^i = \sum_{j}\frac{\partial u^i}{\partial \tilde{u}^j} \tilde{X}^j. $ Consider a change of variables $\phi: \tilde{U}\to U$ and let $\tilde{u}^j$ and $u^i$ be the coordinates respectively on $\tilde{U}$ and $U$.

Question: Let $\tilde{X}$ be a vector field on $\tilde{U}$ with coordinates $\tilde{X}^j$. Show that the coordinates of $X = (d\phi)(\tilde{X})$ are
$$
            X^i = \sum_{j}\frac{\partial u^i}{\partial \tilde{u}^j} \tilde{X}^j.
        $$

The given solution is the following:

We have
$$
        (\underline{d\phi})_{ij} = \frac{\partial u^i}{\partial \tilde{u}^j},
    $$
so that
\begin{align*}
        X^i &= ((d\phi)(\tilde{X}))^i \\ 
            &= \left(\sum_{j} (d\phi)\left(\frac{\partial}{\partial \tilde{u}^j}\right) 
                \tilde{X}^j\right)^i \tag{$1$} \\  
            &= \left(\sum_{j} \frac{\partial u}{\partial \tilde{u}^j} \tilde{X}^j\right)^i 
               \tag{$2$}\\ 
            &= \sum_{j} \frac{\partial u^i}{\partial \tilde{u}^j} \tilde{X}^j.
    \end{align*}

I'm having a really hard time understanding the working done to go from $(1)$ to $(2)$. I know that $d\phi$ is simply the jacobian matrix, so I'm very confused on why it all of a sudden collapses to simply being $u$ in line $(2)$. Any illumination would be helpful, thank you!
 A: I will assume that $\dim \tilde{U}=m$ and $\dim U=n$. Now, in the coordinates that you have, the computation from (1) to (2) is:
$$d\phi \left( \frac{\partial}{\partial \tilde{u}^{j}} \right)=\begin{pmatrix} \frac{\partial u^{1}}{\partial \tilde{u}^{1}} & \frac{\partial u^{1}}{\partial \tilde{u}^{2}} & \cdots & \frac{\partial u^{1}}{\partial \tilde{u}^{m}} \\
\frac{\partial u^{2}}{\partial \tilde{u}^{1}} & \frac{\partial u^{2}}{\partial \tilde{u}^{2}} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots \\
\frac{\partial u^{n}}{\partial \tilde{u}^{1}} & \frac{\partial u^{n}}{\partial \tilde{u}^{2}} & \cdots & \frac{\partial u^{n}}{\partial \tilde{u}^{m}} \end{pmatrix} 
 \begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0  \end{pmatrix}=\begin{pmatrix}  \frac{\partial u^{1}}{\partial \tilde{u}^{j}} \\ \frac{\partial u^{2}}{\partial \tilde{u}^{j}} \\ \vdots \\ \frac{\partial u^{n}}{\partial \tilde{u}^{1}} \end{pmatrix}=\frac{\partial u}{\partial \tilde{u}^{j}},$$
where the column vector has all its entries equal to zero, except for the one in position $j$ (which I will call $e_{j}$). The first equality is how you compute the differential in coordintes: as you said, $d\phi$ is the Jacobian matrix; and the vector $\frac{\partial}{\partial \tilde{u}^{j}}$ is just $e_{j}$ because is one of the vectors of the basis for the tangent space $T_{p}\tilde{U}$. When you work in coordinates, you have to look for the coefficients of your vector in the basis, and to this vectors of coefficients, you apply the Jacobian matrix.
