We are working in ZF.
Problem (part 1). Suppose we have well-ordered set $(M, <)$. How to show that there exists a bijection $f$ between this set and some ordinal $x$ that preserves order?
I can use the following theorem:
Theorem. Let $F$ be a functional and $(A, <)$ be a well-ordered set. Then $\exists!f\ Dom(f) = A$ and $f(x) = F(f|_{\{y \in A \mid y < x\}})$ for all $x \in A$.
If we will take $F := S(x) = x \cup \{x\}$ then we will automatically gain $f$ and for sure it preserves order, but how to rigorously show this fact?
Problem (part 2). How to show that this ordinal $x$ and bijection are unique?
As for uniqueness of the bijection it follows from the theorem, but how to deal with set $x$?
Your $F$ does not work. Note that $F$ should map functions defined on subsets of $A$ to elemens of the desired range (i.e., ordinals), but you define $F$ on ... ordinary elements, so to speak.
Instead, let $F(g):=\operatorname{img}(g)$. Then for example $f(\min A)=F(f|_\emptyset)=\operatorname{img}(f|\emptyset)=\emptyset$. Show by induction that $f(x)$ is the smallest ordinal exceeding all $f(y)$, $y<x$.
For uniqeness, you can get rid of general well-ordered sets. Show that an order-preserving bijection between two ordinals is necessarily the identity.