$U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^l$ be open subsets. Let $f: V \to U$ to smooth. Use $x_1, \dots, x_k$ for the standard coordinate functions on $\mathbb{R}^k$ and $y_1, \dots, y_l$ on $\mathbb{R}^l$.
Consider a tangent vector $Y \in T_yU$ such that $$Y = \sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}.$$
So to me, $Y^j$ is the scalor of a vector in the tangent space of the domain of $f$, so I can move it around? Like
$$(f^*dx_i)\left(Y^j\frac{\partial}{\partial y^j}\right) = Y^j(f^*dx_i)\left(\frac{\partial}{\partial y^j}\right).$$
And same here since $dy_jY$ is just $Y^j$:
$$(dy_j Y)\left(\frac{\partial f_i}{\partial y_j}\right) = \left(\frac{\partial f_i}{\partial y_j}\right) (dy_j Y).$$
In a word, "yes". A fancy way to say this is "tensors are linear over smooth functions". :)