Step 1: The localization sequence. From the localization sequence for the Dedekind domain $\mathbb{Z}$ with fraction field $\mathbb{Q}$:

$K_2(\mathbb{Z}) \hookrightarrow K_2(\mathbb{Q}) \xrightarrow{\oplus \partial_p} \bigoplus_p K_1(\mathbb{F}_p) = \bigoplus_p \mathbb{F}_p^{\times} \xrightarrow{\beta} K_0(\mathbb{Z})$

The map $\beta$ sends $\bar{u} \in \mathbb{F}_p^{\times}$ to $[\mathbb{F}_p] \in K_0(\mathbb{Z})$, but in the localization sequence $\beta$ is actually the zero map because each $[\mathbb{F}_p] - [0] = [\mathbb{Z}/p] \in K_0(\text{torsion modules})$ maps to $0$ in $K_0(\mathbb{Z})$ after the identification with $\widetilde{K}_0$ (which is trivial for $\mathbb{Z}$).

Wait, let us be more careful. The localization sequence is:

$K_2(\mathbb{Z}) \to K_2(\mathbb{Q}) \xrightarrow{\oplus \partial_p} \bigoplus_p \mathbb{F}_p^{\times} \to K_1(\mathbb{Z}) \to K_1(\mathbb{Q})$

The map $\bigoplus_p \mathbb{F}_p^{\times} \to K_1(\mathbb{Z}) = \{\pm 1\}$ sends $\bar{u} \in \mathbb{F}_p^{\times}$ to $\det$ of a lift of $u$ to $\mathbb{Z}^{\times}$, composed with the natural map. But the map $K_1(\mathbb{Z}) = \{\pm 1\} \to K_1(\mathbb{Q}) = \mathbb{Q}^{\times}$ is injective. So the map $\bigoplus \mathbb{F}_p^{\times} \to K_1(\mathbb{Z})$ is actually zero (since any element in the image maps to $1$ in $K_1(\mathbb{Q})$ because the composite $K_2(\mathbb{Q}) \to \bigoplus \mathbb{F}_p^{\times} \to K_1(\mathbb{Z}) \to K_1(\mathbb{Q})$ is zero by exactness).

Therefore: $\oplus \partial_p: K_2(\mathbb{Q}) \to \bigoplus_p \mathbb{F}_p^{\times}$ is surjective and $K_2(\mathbb{Z}) = \ker(\oplus \partial_p)$.

Step 2: Surjectivity of the tame symbol. For each prime $p$ and $u \in \{1, \ldots, p-1\}$, the symbol $\{p, u\}$ satisfies $\partial_p(\{p, u\}) = \bar{u} \in \mathbb{F}_p^{\times}$ and $\partial_q(\{p, u\}) = 1$ for $q \neq p$ (since both $p$ and $u$ are $q$-adic units when $q \nmid pu$, and $u < p$ ensures $\gcd(u, p) = 1$). For primes $q | u$, one must use the bimultiplicativity to adjust. In any case, the tame symbols surject onto each $\mathbb{F}_p^{\times}$.

Step 3: Computation of $K_2(\mathbb{Z})$. We know $\{-1, -1\} \in K_2(\mathbb{Z})$ since $\partial_p(\{-1, -1\}) = (-1)^{1 \cdot 1} (-1)^1/(-1)^1 = (-1) \cdot 1 = -1 \cdot (-1)/(-1) = 1$ (actually computing: $v_p(-1) = 0$ for all $p$, so $\partial_p(\{-1, -1\}) = (-1)^{0 \cdot 0} = 1$). Indeed $\{-1, -1\}$ lies in $K_2(\mathbb{Z}) = \ker(\oplus \partial_p)$ since all tame symbols vanish on it.

To show $K_2(\mathbb{Z}) = \{1, \{-1,-1\}\} \cong \mathbb{Z}/2$, one uses the Hilbert symbol at the real place: $(-1, -1)_\infty = -1$, showing $\{-1,-1\} \neq 1$. The claim that $K_2(\mathbb{Z})$ has no other elements requires showing that any element in $\ker(\oplus \partial_p)$ is $1$ or $\{-1, -1\}$.

This uses Tate's argument: an element $\alpha \in K_2(\mathbb{Q})$ with $\partial_p(\alpha) = 1$ for all $p$ is determined by its images under the Hilbert symbols $(-, -)_p$ at all places. By the product formula $\prod_v (a, b)_v = 1$ and the fact that $(-, -)_\infty$ is the only non-trivial Hilbert symbol for elements of $K_2(\mathbb{Z})$, one concludes $K_2(\mathbb{Z}) \cong \{(-, -)_\infty = \pm 1\} = \mathbb{Z}/2$.

Step 4: The splitting. The exact sequence $0 \to K_2(\mathbb{Z}) \to K_2(\mathbb{Q}) \to \bigoplus_p \mathbb{F}_p^{\times} \to 0$ splits since $\bigoplus_p \mathbb{F}_p^{\times}$ is a direct sum of cyclic groups (hence free as a $\mathbb{Z}$-module modulo the torsion, and each $\mathbb{F}_p^{\times}$ is cyclic). Actually, the splitting comes from the section $\mathbb{F}_p^{\times} \to K_2(\mathbb{Q})$ given by $\bar{u} \mapsto \{p, \tilde{u}\}$ where $\tilde{u} \in \mathbb{Z}$ is a lift. $\square$