The idea of the Yoneda lemma revolves around presheaf categories and "representable presheaves" of an object $X$ denoted by $\operatorname{Hom}(-, X)$.
Note that the functor
$$ Y\colon C \to \operatorname{Hom}(C^{\mathrm{op}}, \mathrm{Set}) $$ $$ X \mapsto \operatorname{Hom}(-, X) $$
is often called the Yoneda embedding.
The presheaf category of a category $C$ is the image of the Yoneda embedding, or the functor category $\operatorname{Hom}(C^{\mathrm{op}}, \mathrm{Set})$
Representable presheaves make up part of this category. The Yoneda embedding takes all objects $X \in C$ and sends them to their representable presheaves $\operatorname{Hom}(-, X)$. These presheaves represent all morphisms into $X$ (and, by the Yoneda lemma, as we'll see, thus $X$ itself).
In a sense these representable presheaves are "building blocks" of the presheaf category because every other presheaf is a colimit of representable presheaves.
In fact morphisms between representable presheaves (natural transformations, presheaf homomorphisms) are in direct correspondence to all morphisms in $C$. This is why the Yoneda embedding is called an embedding. It embeds $C$ into the presheaf category, without losing, adding, or mixing up objects or morphisms.
This is made precise by the following corollary:
Corollary. The Yoneda lemma implies that the Yoneda embedding is full and faithful, meaning there is a bijection between hom-sets
$$ \operatorname{Hom}(X, Y) \cong \operatorname{Hom}(\operatorname{Hom}(-, X), \operatorname{Hom}(-, Y)) $$
You can get this by assigning $P = \operatorname{Hom}(-, Y)$ in the Yoneda lemma.
This is the naturality square for a presheaf homomorphism $\alpha\colon \operatorname{Hom}(-, X) \Rightarrow \operatorname{Hom}(-, Y)$:
$$\begin{CD} \operatorname{Hom}(A, X) @>\alpha_A>> \operatorname{Hom}(A, Y) \\ @VVV @VVV \\ \operatorname{Hom}(B, X) @>>\alpha_B> \operatorname{Hom}(B, Y) \end{CD}$$
If we isolate the top, rename it to $Z$ just for aesthetics
$$\begin{CD} \operatorname{Hom}(Z, X) @>\alpha_Z>> \operatorname{Hom}(Z, Y) \end{CD}$$
We can inspect the diagram
$$\begin{CD} Z @= Z \\ @VVfV @VVhV \\ X @>>{\Box g}> Y \end{CD}$$
This means that given a morphism $f\colon Z \to X$, the only way to form a $h\colon Z \to Y$ is to encode a morphism $g\colon X \to Y$ into $\alpha_Z$. That way $\alpha_Z$ gets associated with that particular $g$. There are no other options. Pretty simple.
So we know what morphisms from representable presheaves to other representable presheaves look like, but what do morphisms from representable presheaves to any arbitrary presheaf look like?
Well the Yoneda lemma states that the morphisms from a presheaf representing $X$ to any other presheaf $P$ are in a bijective relationship with the set $P(X)$.
Formally, the statement is
$$ P(X) \simeq \operatorname{Hom}(\operatorname{Hom}(-, X), P) $$
The natural transformation expressed has the signature $\alpha_Z\colon \operatorname{Hom}(Z, X) \to P(Z)$
So, by the same logic, given an $f\colon Z \to X$, the Yoneda lemma states that the only way to construct a $P(Z)$ is to lift $f$ into $P(f)\colon P(X) \to P(Z)$ and then encode a $x \in P(X)$ into $\alpha_Z$. There are no other options.
You might be thinking, why can't we just assume a $P(Z)$ and we don't have to use $f$? Yes, the principle is correct, just set $Z = X$ and the same logic applies. In this case, identity is a choice for $f$.