# Why are interior products necessary when 1-forms are already dual to vectors?

I am reading a book about differential geometry that introduces an operator $$i$$, called the "interior product," that takes vectors and produces something that can act on 1-forms. Their rules are, $$i\left(\frac{\partial}{\partial x_i}\right)\mathrm{dx_j} = \delta_{ij}$$ and, $$i\left(\frac{\partial}{\partial x_i}\right)f = 0$$ Now, the second rule is unique, and is definitely not what the vector $$\partial/\partial x$$ would do normally. Still, I don't understand why the first one should apply. Vectors are already dual to one-forms, so why do we need a map $$i$$ that takes vectors to something that can act on one-forms?

This is especially bothersome to me, because interior products usually allow you to measure vectors against other vectors in the same space. I am not sure how to interpret these laws, which apparently describe an "inner product" between spaces that already have a bilinear function to be dual under!

No, the point is that interior product maps $$(k+1)$$-forms to $$k$$-forms (for all $$k\ge 0$$). This has nothing to do with inner products. Indeed, it's the process of undoing the wedge product (so-called "adjoint operation"), which sends $$k$$-forms to $$(k+1)$$-forms.
It's based on contraction (evaluating a $$1$$-form on a vector), and that, indeed, is the rule they gave you for the case $$k=0$$.
More generally, if we have a vector $$v$$ and, for example, two $$1$$-forms $$\omega$$ and $$\eta$$, then $$\iota_v(\omega) = \omega(v) \quad\text{and}\quad \iota_v{\eta} = \eta(v),$$ and then $$\iota_v(\omega\wedge\eta) = \omega(v)\eta - \eta(v)\omega.$$ (I'll leave it to you to figure out why the negative sign is there.)
An important application of this notion is the following: If you have an oriented surface $$S$$ in $$\Bbb R^3$$, with outward pointing unit normal $$\vec n$$, then you get the area $$2$$-form on $$S$$ (written $$dS$$ or $$d\sigma$$ in calculus books) by taking $${"}dS{"} = \iota_{\vec n} (dx_1\wedge dx_2\wedge dx_3).$$ This generalizes to oriented hypersurfaces in any dimension.