Why does doubling a covector make the "ruler markings" closer together? I was watching this video about how to visualize covectors as "ruler markings" or "topographic maps" with which to measure vectors. If $\alpha$ is a covector and $v$ is a vector, my confusion starts when we ask what $(2\alpha)(v)$ is and how to visualize it. I understand that $(2\alpha)(v) = 2(\alpha(v))$ and that the ruler markings get denser (so that the vector $v$ pierces more lines). But we could also have defined things so that $(2\alpha)(v) = \frac12(\alpha(v))$ and where the ruler markings get farther apart.
Here's another way I am thinking about this. We have objects in the real world whose lengths we can measure, and we have units with which to measure these objects. If "mouse" is an object and "cm" is a unit, we can write (mouse,cm) = 8 to mean that the mouse has length 8 cm. Now we can have different units, and in particular we want to do some arithmetic with units and define e.g. what "2cm" means as a unit. We could write "2cm" to mean "measure the object using units twice as long as cm" so that (mouse,2cm) = 4, or we could write "2cm" to mean "twice the number you get if you measure the object using cm" so that (mouse,2cm) = 16. As far as I know, this is just a convention. The convention has the nice property that $(\cdot,\cdot)$ is linear in each argument. (I am using this awkward notation because the units here are like inner products and we can think of this as asking whether $\langle v,2u\rangle$ should equal $2\langle v,u\rangle$ or $\frac12\langle v,u\rangle$.)
So is there a deeper reason (than just "inner products/covectors should be linear") for defining things this way?
 A: I think that by using an inner product, you have introduced too many rulers to the party, since an inner product already defines its own notion of length: $\lVert v \rVert := \sqrt{(v, v)}$. Furthermore, there is no reason why $(\text{mouse}, \text{cm})$ is the correct way of measuring the length of a mouse in centimetres (and in fact, it is not).
Let's say I have a one-dimensional vector space $L$. I have a yardstick $y \in L$ which I wish to use to measure a mouse $m \in L$. The length of the mouse $\ell \in \mathbb{R}$ in the solution to the equation $\ell y = m$. You can see here clearly why doubling my yardstick halves the length, since if $\ell y = m$ then $\frac{\ell}{2} (2y) = m$ also. You can also see that $\ell$ has a linear relationship to $m$, for example $(2 \ell) y = 2m$. Measuring is linear, but changing the side of the yardstick is not linear (it is the reciprocal).
Getting more abstract, to any nonzero yardstick $y \in L$ we can define the measure-with-respect-to-$y$ functional $\ell_y \colon L \to \mathbb{R}$, which is the unique functional such that $\ell_y(m) y = m$ for all $m \in L$. (A simpler definition is the unique functional such that $\ell_y(y) = 1$). The question now is: how is $\ell_{2y}$ related to $\ell_y$? We can find that
$$ 1 = \ell_{2y}(2y) = (2 \ell_{2y})(y)$$
which shows that $2 \ell_{2y} = \ell_y$, or that $\ell_{2y} = \ell_y / 2$. So the assignment $y \mapsto \ell_y$ is nonlinear. This is important: the functional $2 \ell_y$ corresponds to the yardstick $y/2$, not to $2y$.
Finally, what happens if $L$ already comes with an inner product? There must exist a vector $z_y \in L$ such that $\ell_y(m) = (m, z_y)$ for all $m \in L$. It is not hard to see that $z_y = \frac{y}{(y, y)}$ works, since then we have $(y, z_y) = 1$ as required. Again, we can see that $z_{2y} = z_y / 2$. You might also recognise $(m, z_y)$ as the formula for the orthogonal projection of $m$ onto $y$.
Long answer short the length of a mouse in centimetres is $\frac{(\text{mouse}, \text{cm})}{(\text{cm}, \text{cm})}$, not $(\text{mouse}, \text{cm})$, and converting yardsticks to rulers works like reciprocals, not like a linear map.
