A book i'm reading says consider the "dictionary" $x = u + a$ and $y = v + b$ between the $x,y$ and $u,v$ coordinate systems, where the $u,v$ coordinate axes are obtained by translating the $x,y$ $a$ units to the right, and $b$ units upward.
It makes sense geometrically that given the $(x,y)$ coordinates of a point $P$ of the plane, the "dictionary" above allows us to find the $(u,v)$ coordinates of the same point, and vice versa.
Then the book wants to illustrate that sometimes the origin and axes direction remain unchanged but the unit of length may differ. And so they say suppose we have the following "dictionary" between $x,y$ and $u,v$ planes:
$u = 2x, v = 3y$. Where they depict the point $(1,0)$ in the $u,v$ plane being in the position of the point $(2,0)$ in the $x,y$ plane. So the step sizes in the $u,v$ plane are larger.
After many hours I think I figured out that the use of the word dictionary here (unlike the first example for some mysterious reason?) is misleading.
The transformation above does not allow us to feed in $(x,y)$ coordinates of a point $P$ and determine the $(u,v)$ coordinates of the same point. It seems to instead tell us how to obtain the $u-v$ plane basis from the $x-y$ one. And it seems that the dictionary between the two system is given by $u = \frac{1}{2}x$, and $v = \frac{1}{3}y$.
Is my interpretation correct? Would you agree that the description of dictionary for the second example (but not the first) was misleading? Since it seemed to give a transformation, and not the change of basis formulas.
What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.
I will use Einstein summation notation, let me know if it's confusing for you. I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.
A linear transformation tells us how to find the new basis vectors from the old basis vectors: $$ \tilde{e}_i = A_i^j e_j $$ In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases: $$ v = \tilde{v^i}\tilde{e_i} = v^j e_j $$ How do we get the new vector components $\tilde{v^i}$? $$ \tilde{v^i}\tilde{e_i} = \tilde{v^i}A_i^j e_j $$ That is, by comparing the two forms: $$ \tilde{v^i}A_i^j = v^j $$ This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$. $$ \tilde{v^i} = v^j (A^{-1})_j^i$$
So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.