I understand the 1-Yoneda lemma well, and I know how to use it in proofs, but I have trouble understanding how the 2-Yoneda lemma exactly works.
Say for simplicitly I have a strict 2-category $K$ and two 1-cells $l:A\to B$ and $r:B\to A$. Say I manage to show that for each other 0-cell $X$ and test-1-cells $a:X\to A$ and $b:X\to B$ there is a bijection of 2-cells
$$K(la,b) = K(a,rb)$$
which varies 2-naturally in $a$ and $b$. How can I use the 2-Yoneda lemma to conclude that $l$ and $r$ must be adjoint to each other in $K$? If you answer, please give me a completly rigorous argument with no steps missing.
If I understand correctly, then the embedding $y:K \to 2Cat(K^{op},Cat)$ is hom-cat-wise an equivalence, so it is enough to show that $y(l)$ and $y(r)$ are adjoint to each other in the 2-category of 2-functors, transformations and modifications. How do I get this from $K(la,b)=K(a,rb)$?
Remark. Lack's "2-categories companion" claims that it is enough to find a 2-natural bijection $K(la,b) = K(a,rb)$, but does not really explain why.
Just to fix terminology. You have a strict 2-category $\mathcal{K}$. There is a strict 2-category $\operatorname{Fun}(\mathcal{K}^\text{op},\mathsf{Cat})$ of strict 2-functors, 2-natural transformations and modifications. The 2-Yoneda embedding $$Y_\mathcal{K}:\mathcal{K} \rightarrow \operatorname{Fun}(\mathcal{K}^\text{op},\mathsf{Cat}), x \mapsto \mathcal{K}(-,x)$$ is a strict 2-functor, which is locally an isomorphism of Hom-categories.
Suppose you have morphisms $l:x\rightarrow y$ and $r:y\rightarrow x$ in $\mathcal{K}$, which induce an adjunction in the 2-category $\operatorname{Fun}(\mathcal{K}^\text{op},\mathsf{Cat})$ under the Yoneda embedding. Explicitly this means that the induced 2-natural transformations $Yl=l_\ast:\mathcal{K}(-,x) \Rightarrow \mathcal{K}(-,y)$ and $Yr=r_\ast:\mathcal{K}(-,y) \Rightarrow \mathcal{K}(-,x)$ admit a unit and counit, i.e. modifications $\eta_\ast:1_{\mathcal{K}(-,x)}\Rrightarrow r_\ast l_\ast$ and $\epsilon_\ast:l_\ast r_\ast \Rrightarrow 1_{\mathcal{K}(-,y)}$ satisfying the triangle identities. The 2-Yoneda embedding having isomorphisms of Hom-categories implies that $\eta_\ast = Y_\mathcal{K}(\eta)$ and $\epsilon_\ast = Y_\mathcal{K}(\epsilon)$ for some 2-cells $\eta,\epsilon$ in $\mathcal{K}$, which satisfy the triangle identities in $\mathcal{K}$.
This is one way to use 2-Yoneda to get an adjunction in $\mathcal{K}$. Unfortunately, to me the criterion you cite doesn't make sense at all for an arbitrary 2-category $\mathcal{K}$. What is "2-natural bijection $\mathcal{K}(la,b) \cong \mathcal{K}(a,rb)$" even supposed to mean?
What is true, is that you have a Hom-wise criterion for adjunctions between 2-categories. In this setting we assume we have 2-functors $\mathcal{L}:\mathcal{A} \rightarrow \mathcal{B}$ and $\mathcal{R}:\mathcal{B} \rightarrow \mathcal{A}$. To say that they are adjoint amounts to giving a 2-natural isomorphism $\varphi:\mathcal{B}(\mathcal{L}-,-) \cong \mathcal{A}(-,\mathcal{R}-)$, i.e. an isomorphism of 2-profunctors $\mathcal{A}^\text{op}\times\mathcal{B} \rightarrow \mathsf{Cat}$. Precomposing with $\mathcal{L}$ and using that $\mathcal{L}$ is a 2-functor gives us a 2-natural transformation $\widetilde{\eta}:\mathcal{A}(-,-)\Rightarrow\mathcal{B(L-,L-)} \cong \mathcal{A(-,RL-)}:\mathcal{A^\text{op}\times A}\rightarrow\mathsf{Cat}$. Using the fact that $\mathsf{2Cat}$ is a 2-cartesian closed 2-category, this amounts to a 2-natural transformation $\widehat{\eta}$ between the 2-functors $X \mapsto \mathcal{A}(-,X) = Y_\mathcal{A}(X)$ and $X \mapsto \mathcal{A}(-,\mathcal{RL}X) = Y_\mathcal{A}(\mathcal{RL}X)$ in $\operatorname{Fun}(\mathcal{A},\operatorname{Fun}(\mathcal{A}^\text{op},\mathsf{Cat}))$. By 2-Yoneda this is realized by a collection of morphisms $(\eta_X:X\rightarrow \mathcal{RL}X)_{X\in\mathcal{A}}$. The fact that $\widehat{\eta}$ is 2-natural in $X$ translates into the fact that $\eta:1_\mathcal{A}\Rightarrow\mathcal{RL}$ is a 2-natural transformation. In a completely analogous way we obtain the desired 2-natural transformation $\epsilon:\mathcal{LR}\Rightarrow 1_\mathcal{B}$. Now by definition $\eta_X = (\eta_X)_\ast(1_X) = \widetilde{\eta}_{X,X}(1_X) = \varphi_{X,\mathcal{L}X}(1_{\mathcal{L}X})$ and by the same reasoning $\epsilon_X = \varphi^{-1}_{\mathcal{R}X,X}(1_{\mathcal{R}X})$. For any morphism $f\in\mathcal{B}(\mathcal{L}X,Y)$ the commutativity of the diagram $$\begin{array}{ccc} \mathcal{B}(\mathcal{L}X,\mathcal{L}X) & \overset{\phi_{X,\mathcal{L}X}}{\longrightarrow} & \mathcal{A}(X,\mathcal{RL}X)\\ \downarrow f_\ast && \downarrow (\mathcal{R}f)_\ast\\ \mathcal{B}(\mathcal{L}X,Y) & \overset{\phi_{X,Y}}{\longrightarrow} & \mathcal{A}(X,\mathcal{R}Y) \end{array}$$ implies that $\phi_{X,Y}(f) = \mathcal{R}f \circ \eta_X$. Similarly we get that for $g\in\mathcal{A}(X,\mathcal{R}Y)$ we have $\phi^{-1}_{X,Y}(g) = \epsilon_Y \circ \mathcal{L}(g)$. With this explicit formula we can deduce the triangle equalities.
Note: This proof almost verbatim applies in general enriched category theory.