Suppose we have derived
$\phi^*(\beta)=\sum_{jR}b_R(\partial y^R/\partial x^j)dx^j$
as here in (2.24). How do I obtain the formula (2.23) from (2.24). I.e. I want substitute $v$ into (2.24) and get $\phi^*(\beta)(v)$; what should be this last expression in coordinates?
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Write $v$ as $$\textstyle v = \sum_j v^j \frac{\partial}{\partial x^j} $$ and multiply it from the left by $$\textstyle \phi^*(\beta)=\sum_{jR} b_R \frac{\partial y^R}{\partial x^j}dx^j. $$ You will see that this is the same as the product of $\phi_*$ with $v$, that is $$\textstyle \sum_R \left(\sum_j \frac{\partial y^R}{\partial x^j} v^j \right) \frac{\partial}{\partial y^R}, $$ multiplied by $$\textstyle \beta = \sum_R b_R dy^R $$ from the left.
In terms of matrices, this is indeed the matrix multiplication $$ \begin{pmatrix} b^1 & \dots & b^r\end{pmatrix} \begin{pmatrix} \frac{\partial y^1}{\partial x^1} & \dots & \frac{\partial y^1}{\partial x^n} \\ \vdots & & \vdots \\ \frac{\partial y^r}{\partial x^1} & \dots & \frac{\partial y^r}{\partial x^n} \end{pmatrix} \begin{pmatrix} v^1 \\ \vdots \\ v^n \end{pmatrix} $$ so $\phi^*(\beta)$ is just the product of the row vector on the left with the Jacobian matrix of $\phi_*$, everything relative to the given coordinates, of course.