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Yes, the interior derivative is always specified relative to a vector field. It is very simple. In fact, you quote a complete definition in your question.
Then the interior derivative $ι_vω$, which is the n−1-form defined by
$ι_vω(X_2,…,X_n)=ω(v,X_2,…,X_n)$
That's really all there is to it. You take some n-form (with n>0) and to figure out how it acts on some other set of vector fields, you simply "insert" the vector field you took the interior product with into the arguments. Look at the above expression, it's simpler than words can describe. It shows how a n-1 form $ι_vω$ acts on a series of vector fields $X_2,…,X_n$.
Note, however, that this doesn't work on 0-forms. You can see this one of two ways. First, it takes n forms to n-1 forms, and there's no such thing as -1 forms. Second, if you look at how its defined above, there's nowhere to "insert" the vector field as an argument, because a 0-form takes 0 arguments. So the interior derivative of a 0-form (a function) is just always 0.
2.
As Dylan Pointed out, the interior product is the same as the interior derivative. See the Wikipedia article.
3.
The exterior product is not the same as the exterior derivative. The exterior product is just another name for the wedge product. The exterior derivative is the $d$ operator.
There is an close relationship between exterior derivatives, interior derivatives, and lie derivatives, see Cartan's identity:
$\mathcal L_X\omega = \mathrm d (\iota_X \omega) + \iota_X \mathrm d\omega$
It's useful to think of differential forms as antisymmetric multilinear mappings, i.e a $p$-form eats $p$ arbitrary vector fields and gives you an ordinary function of $n$ variables (where $n$ is the dimension of your manifold).
The interior product is a mapping from a $p$ form $\omega$ to a $(p-1)$ form, since you've fixed one argument of the $p$ form to be a particular vector field, say $X$ and as a result, $\iota_X \omega$ can only act on $(p-1)$ vector fields now.
To get the explicit formula in local coordinates for the interior product of a form with a vector field, you can use the formula for two forms $$(\iota_X \eta)_i = X^j\eta_{ji}$$(summation convention) to get $\iota_X d\omega$:
$(i_X d\omega) = X^j (d\omega)_{ji} dx^{i}.$
If this is still confusing, consider working your example out directly by the definition of everything:
$d\omega = 3dx\wedge dy - 14zxdx\wedge dz$, which written in terms of tensors is given by
$d\omega = 3(dx\otimes dy - dy \otimes dx) - 14zx (dx\otimes dz - dz\otimes dx)$
So then contracting in the first argument with the vector field X gives you:
$\iota_X d\omega = 3(dx(X)\otimes dy - dy(X) \otimes dx) - 14zx (dx(X)\otimes dz - dz(X)\otimes dx)$
$\iota_X d\omega = 3[dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dy - dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) \otimes dx] - 14zx [dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dz - dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dx]$
Now since $dx$ is a linear functional (field) that are dual to $\partial_x$ (and so on for $y$ and $z$), we get that
$dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = y$
$dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 2z$
$dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 3xy$.
Using this, gives you
$\iota_X d\omega = [-3(2z)+ 14zx(3xy)]dx + (3y)dy - (14zxy)dz$
as your one-form.
Note that this is a laborious way and only serves to illustrate the definitions to you. In practice, it would be faster to get the answer by the formula given above.
Now you can take any arbitrary vector field $Y$, do the same contraction with this now one-form to get a function of $3$ variables.
Best Answer
$(f^*\omega)_y(Y_1,..,Y_k)=\omega_{f(y)}(df_y(Y_1),...,df_y(Y_k))$
$i_X\omega_x(X_1,..,X_{k-1})=\omega_x(X,X_1,..,X_{k-1})$ we deduce that
$(f^*i_X\omega)_y(Y_1,..,Y_{k-1})=\omega_{f(y)}(X(f(y)),df_yX_1(y),..,df_yX_{k-1}(y))$
$(f^*X)(y)=df^{-1}_{f(y)}X(f(y))$, we deduce that $(i_{f^*X}f^*\omega)_y(Y_1,..Y_{k-1})=$
$(f^*\omega)_y(df_{f(y)}^{-1}(X(f(y)),Y_1,..,Y_{k-1}))$=
$\omega_{f(y)}(df_y(df_{f(y)}^{-1}(X(f(y))),df_yY_1,..,df_y.Y_{k-1})=$
$\omega_{f(y)}(X(f(y)),df_yX_1(y),..,df_yX_{k-1}(y))$.