I am reading this book, GAUGE FIELDS, KNOTS AND GRAVITY by John Baez and J P Muniain. The author uses the notation $vf$ for the directional derivative along the vector $v$ of the smooth function $f$ on a manifold $M$. The authors later define a vector field(p.25) as
Now, a vector field $v$ on $M$ is defined to be a function from $C^{\infty}(M)$ to $C^{\infty}(M)$ satisfying the following properties: $$ \begin{aligned} v(f+g) & =v(f)+v(g) \\ v(\alpha f) & =\alpha v(f) \\ v(f g) & =v(f) g+f v(g), \end{aligned} $$ for all $f, g \in C^{\infty}(M)$ and $\alpha \in \mathbb{R}$.
the vector field defined here is $v$, which has earlier been defined to be the directional derivative. Going a few pages ahead, on page 41, after defining a 1-form, he writes an example for it which is
$$df(v)=vf$$
So my question is, how is the second case, which is the result of taking the directional derivative of the function $f$, which is another smooth function, become a 1-form and the same thing in the earlier case, although as something which is yet to act on a function, is a vector field.
The relation becomes transparent when working with local coordinates $p = (x_1,\ldots,x_n) \in M$. Indeed, then you have : $$ \begin{cases} \displaystyle v(p) = \sum_{i=1}^n v_i\frac{\partial}{\partial x_i} \in T_pM \\ \displaystyle \mathrm{d}f(p) = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\mathrm{d}x_i \in T_p^*M \end{cases} $$ hence $$ \mathrm{d}f(p)(v) = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\mathrm{d}x_i \left(\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\right) = \sum_{i,j=1}^n v_j\frac{\partial f}{\partial x_i}\mathrm{d}x_i \left(\frac{\partial}{\partial x_j}\right) = \sum_{i=1}^n v_i\frac{\partial f}{\partial x_i} = v(p)(f) $$ because $\mathrm{d}x_i\left(\frac{\partial}{\partial x_j}\right) = \delta_{ij}$, since $\{\mathrm{d}x_i\}_{1\le i\le n}$ is the dual basis of $\left\{\frac{\partial}{\partial x_j}\right\}_{1\le j\le n}$.