TL;DR: In your chain of equalities, the last $=$ does not hold in general (the Lipschitz norm is defined as a supremum). As a sanity check, consider the linear isomorphism $f:\mathbb{R}^2\to\mathbb{R}^2, (x,y)\mapsto (2x,3y)$. $f$ is bi-Lipschitz and its distortion is $3/2$. However the error points to some other notion (what I call below the "Lipschitz coconstant"), which can be used to write a relationship between the distortion and constants of Lipschitz type for any function. Also note that the formula (in the form given) Matoušek gives applies to any bijection, not only to bi-Lipschitz bijections.
Two discussions that are relevant are Properties of conorm definition, Motivation behind the definition of Banach-Mazur Distance the discussion at hand can be considered as the Lipschitz analog of the linear theory.
Below I present some definitions and observations that hopefully will clarify the distortion formula.
Let $X$ and $Y$ be metric spaces, $f:X\to Y$ be a function, $D\in\mathbb{R}_{\geq1}$. Then $f$ is a $D$-embedding if
$$\exists r\in\mathbb{R}_{>0},\forall x_1,x_2\in X: r d_X(x_1,x_2)\leq d_Y(f(x_1),f(x_2))\leq rD d_X(x_1,x_2).$$
The distortion of functions are defined as follows :
$$\delta: F(X;Y)\to[1,\infty], f\mapsto \inf\{D\in\mathbb{R}_{\geq1}\,|\, f \text{ is a }D\text{-embedding}\}.$$
(Throughout this answer if a set over a supremum or infimum is taken is empty the value is interpreted to be the appropriate boundary value.)
Obs.1: If $f$ is not injective, then it can't be a $D$-embedding for any $D$, whence $\delta(f)=\infty$.
Obs.2: If $f$ is an injective isometry (= preserves distances), then $\delta(f)=1$, but not vice versa, e.g. $\delta(f)=1$ for $f:\mathbb{R}\to\mathbb{R}, t\mapsto 2t$.
Define the Lipschitz constant and Lipschitz coconstant as follows, respectively:
$$\Lambda: F(X,Y)\to[0,\infty],\,\, f\mapsto \sup_{\substack{x_1,x_2\in X\\ x_1\neq x_2}}\dfrac{d_Y(f(x_1),f(x_2))}{d_X(x_1,x_2)}$$
$$\lambda: F(X,Y)\to[0,\infty],\,\, f\mapsto \inf_{\substack{x_1,x_2\in X\\ x_1\neq x_2}}\dfrac{d_Y(f(x_1),f(x_2))}{d_X(x_1,x_2)}$$
It's clear that $0\leq \lambda(f)\leq \Lambda(f)\leq \infty$. Define further Lipschitz bolicity (or Lipschitz condition number) as their ratio:
$$\beta: F(X,Y)\to [1,\infty],\,\,f\mapsto \dfrac{\Lambda(f)}{\lambda(f)}$$
Obs.3: If $X$ and $Y$ are normed vector spaces and $T:X\to Y$ is a (not necessarily continuous) linear map, then $\Lambda(T)=\Vert T\Vert$ is the operator norm, $\lambda(T)=\mathfrak{m}(T)$ is the operator conorm, and $\beta(T)=\operatorname{bol}(T)$ is the bolicity (as discussed in the above links).
Obs.4: $\Lambda(f)<\infty$ iff $f$ is (globally) Lipschitz.
Obs.5: If $0< \lambda(f)$, then $f$ is injective, but not vice versa, e.g. $f:\mathbb{R}\to\mathbb{R}, t\mapsto t^3$ is injective and $\lambda(f)=0$.
Obs.6: If $\beta(f)<\infty$, then $f$ is injective and Lipschitz.
Obs.7: If $f$ is bijective, then $\Lambda(f^{-1})=\dfrac{1}{\lambda(f)}$. (This corresponds to your chain of equalities.)
Proof: Let $y_1,y_2\in Y$. Then
$$\lambda(f) d_X(f^{-1}(y_1),f^{-1}(y_2))\leq d_X(f(f^{-1}(y_1)),f(f^{-1}(y_2))) = d_Y(y_1,y_2)$$
gives $\Lambda(f^{-1})\leq \dfrac{1}{\lambda(f)}$. Let $\epsilon\in\mathbb{R}_{>0}$. Then $\lambda(f)<\frac{1}{\frac{1}{\lambda(f)}-\epsilon}$, whence by the definition of $\lambda(f)$ there are $x_1,x_2\in X$ with $x_1\neq x_2$ such that
$$\lambda(f)\leq \dfrac{d_Y(f(x_1),f(x_2))}{d_X(x_1,x_2)}<\frac{1}{\frac{1}{\lambda(f)}-\epsilon}.$$
Writing $y_1=f(x_1)$ and $y_2=f(x_2)$ establishes the equality.
Obs.8: If $f$ is bijective, then $\beta(f)=\Lambda(f)\Lambda(f^{-1})$, so that we recover Matoušek's formula.
Obs.9: $f$ is bijective and $\beta(f)<\infty$ iff $f$ is a bi-Lipschitz homeomorphism.
Claim: For any function $f: X\to Y$, $\delta(f)=\beta(f)$.
Proof: First suppose $\delta(f)=\infty$. Then $\forall D\in\mathbb{R}_{\geq1},\forall r\in\mathbb{R}_{>0}, \exists x_{D,r},y_{D,r}\in X$ such that
$$\text{either } \lambda(f)\leq \dfrac{d_Y(f(x_{D,r}),f(y_{D,r}))}{d_X(x_{D,r},y_{D,r})}<r\quad \text{ or }\quad rD\leq \dfrac{d_Y(f(x_{D,r}),f(y_{D,r}))}{d_X(x_{D,r},y_{D,r})}\leq \Lambda(f).$$
For any $n\in\mathbb{Z}_{\geq1}$, put $D=n^2$ and $r=\dfrac{1}{n}$. Then either for infinitely many $n$ the first estimate holds, in which case $\lambda(f)=0$, or for infinitely many $n$ the second estimate holds, in which case $\Lambda(f)=\infty$. In either case we thus have $\beta(f)=\infty$.
Conversely suppose $\delta(f)<\infty$. Then there is a $D\in\mathbb{R}_{\geq1}$ and an $r\in\mathbb{R}_{>0}$ such that
$$\forall x_1,x_2\in X: 0<r\leq \dfrac{d_Y(f(x_1),f(x_2))}{d_X(x_1,x_2)}\leq rD< \infty,$$
whence $0<r\leq\lambda(f)\leq \Lambda(f)\leq rD<\infty$, so that $\beta(f)\leq D<\infty$. Thus $\delta(f)=\infty \iff \beta(f)=\infty$. So we may assume that both the distortion and the Lipschitz bolicity of $f$ are finite. We have just proved that for any $D\in\mathbb{R}_{\geq1}$ such that $f$ is a $D$-embedding, $\beta(f)\leq D$, whence by definition $\beta(f)\leq \delta(f)$. Conversely for any $x_1,x_2\in X:$
$$\lambda(f)d_X(x_1,x_2)\leq d_Y(f(x_1),f(x_2))\leq \Lambda(f)d_X(x_1,x_2)=\lambda(f)\beta(f)d_X(x_1,x_2);$$
so that, since $0<\lambda(f)$, $f$ is a $\beta(f)$-embedding, thus $\delta(f)=\beta(f)$.
