Let
$0$-form $f =$ function $f$
$1$-form $\alpha^{1} =$ covariant expression for a vector $\bf{A}$
Then consider the following dictionary of symbolic identifications of expressions expressed in the language of differential forms on a manifold and expressions expressed in the language of vector calculus in $\mathbb{R}^{n}$.
$i_{\bf{v}}(\alpha^{1}) \iff \bf{v}\cdot{\bf{A}}$
$df \iff \text{grad}f$
Now, with $\alpha^{1} = a_{i}dx^{i}$,
why is $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})\cdot{d\bf{x}}$
and not $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})$?
Edit to my answer:
$\frac{d}{dt}\int_{C}{\bf{A}\cdot{dx}}$
$\iff \frac{d}{dt}\int_{C(t)}{a_{1}dx^{1}+a_{2}dx^{2}+a_{3}dx^{3}}$
$= \frac{d}{dt}\int_{C(t)}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \frac{d}{dt}\int_{W(t)}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{W(t)} \mathcal{L}_{X}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{W(t)} \mathcal{L}_{{\bf{v}}+\partial / \partial t}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{W(t)} \mathcal{L}_{{\bf{v}}}\left(\sum\limits_{i}a_{i}dx^{i}\right)+\mathcal{L}_{\partial / \partial t}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{W(t)} \mathcal{L}_{\bf{v}}\left(\sum\limits_{i}a_{i}dx^{i}\right) + \left(\sum\limits_{i}\frac{\partial a^{i}}{\partial t}dx^{i}\right)$
$= \int_{W(t)} \left(\sum\limits_{i}\frac{\partial a^{i}}{\partial t}dx^{i}\right) + i_{\bf{v}}\textbf{d}\left(\sum\limits_{i}a_{i}dx^{i}\right)+\textbf{d}i_{\bf{v}}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{C(t)} \left(\sum\limits_{i}\frac{\partial a^{i}}{\partial t}dx^{i}\right) + i_{\bf{v}}\textbf{d}\left(\sum\limits_{i}a_{i}dx^{i}\right)+\textbf{d}i_{\bf{v}}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$= \int_{C} \left(\sum\limits_{i}\frac{\partial a^{i}}{\partial t}dx^{i}\right) + i_{\bf{v}}\textbf{d}\left(\sum\limits_{i}a_{i}dx^{i}\right)+\textbf{d}i_{\bf{v}}\left(\sum\limits_{i}a_{i}dx^{i}\right)$
$\iff \int_{C}\left[\frac{\partial {\bf{A}}}{\partial t}-{\bf{v}}\times \text{curl}{\bf{A}}+\text{grad}({\bf{v\cdot{A}}})\right]\cdot{d\bf{x}}$
In order for there to be an integration variable in the final line, there must be the integration variable $d{\bf{x}}$, but the integration variable is not supposed to be there, right?
The gradient on $\mathbb{R}^n$ is a vector field and so you need to transform $\operatorname{grad}(\mathbf{v} \cdot \mathbf{A})$ into a one-form.
If $\mathbf{X} = (X_1, \dots, X_n)$ is a vector field on $\mathbb{R}^n$, you can identify it with a one-form $X_1 dx^1 + \dots X_n dx^n$ which you can write as $\mathbf{X} \cdot d \mathbf{x}$ if you insist.