I come from a machine learning background and recently was working on understanding self normalizing networks. It requires me to know the proof of banach fixed-point theorem which needs me to know contraption mapping which got me to its wikipedia page. By now you know i deserve a mathematical layman treatment in this
I get the idea of contraction mapping when the transformation is from $M \rightarrow M$ and the distance metric remains the same, but:
On the wikipedia page of contraption mapping it says
More generally, the idea of a contractive mapping can be defined for maps between metric spaces. Thus, if $(M,d)$ and $(N,d')$ are two metric spaces, then ${\displaystyle f:M\rightarrow N}$ is a contractive mapping if there is a constant ${\displaystyle k<1}$ such that ${d'(f(x),f(y))\leq k\,d(x,y)} $
It eludes me how are the things getting compared for two different metric spaces. what does it even mean when we say $d(.) > d^\prime(.)$ one would think these are two different distance metrics and hence not comparable
To expand on Ryan's comment: A metric $d$ defines the distance between two points. Distances are non-negative real numbers. So while $x, y \in M$, the distance between them, $d(x,y)$ is just a real number.
$f$ is by assumption a map which assigns points in $N$ to points in $M$. So, while $x, y \in M, f(x)$ and $f(y)$ are in $N$. And therefore the metric $d'$ on $N$ can measure the distance between them. And again, that distance $d'(f(x), f(y))$ is just a non-negative real number.
As they are both non-negative real numbers, we can do anything we want to do with them that is defined on non-negative real numbers. We could add them together, we could multiply them, we could raise one to the power of the other. In particular, we can compare one to the other to see which is larger.
Since you came at this while researching the Banach fixed point theorem, be aware that while a map between any two metric spaces can be a contraction, this theorem only applies to maps from a metric space to itself. A map $f$ from $M$ to $N$ may or may not be a contraction, but it can only have a fixed point if $M$ and $N$ intersect (since a fixed point would have to be in both of them). The Banach fixed point theorem only guarantees a contraction $f$ has a fixed point if $f$ maps $M$ to itself, and $M$ is complete (has no holes).