The Mikhlin multiplier states the following:
Let $m : \mathbb{R}^{n} \backslash \{0\} \rightarrow \mathbb{C}$ satisfy the following:
$$|\partial^{\alpha}m(\xi)|\leq C_{0}|\xi|^{-|\alpha|}, \forall \alpha \in \mathbb{N}_{0}^{n} \text{ i.e. alpha is a multi-index with } |\alpha| \leq n+2.$$ $$\text{ Then, for all } 1 < p < \infty, \exists B=B(m,n,p) > 0 \text{ such that } ||T_{m}f||_{L^{p}} \leq B||f||_{L^{p}}, \forall f \in \mathcal{S}(\mathbb{R}^{n}).$$
In proving this one considers a partition of unity as follows $\Psi \in C_{c}^{\infty}(\mathbb{R}^{n}),$ with support of $\Psi \subseteq \mathbb{R}^{n} \backslash \{0\}$ and $ \sum_{j=- \infty}^{\infty} \Psi(\frac{x}{2^{j}}) =1 \,\forall x \neq 0$. This can be choosen radial and non-negative.
Then defining $\Psi_{j}(x):= \Psi(\frac{x}{2^{j}})$ for $\Psi$ as above and observing that each $\Psi_{j}$ is supported in a dyadic annulus of size $2^{j}$ and given $j \in \mathbb{Z}, m_{j}(\xi):=\Psi_{j}(\xi)m(\xi)$
How the The Fourier multiplier of the above $m_{j}(\xi)$ is $m_{j}(D)$ How do we obtain $m(D)f = \sum_{j}m_{j}(D)f$ with convergence in $L^{2}$ (I know we have to use Plancherel and the Dominated Convergence Theorem somehow)
As you mentioned, we can first apply Plancherel's theorem to work in the frequency variable. We now need to show that $$m(\xi) \hat{f}(\xi) = \sum_j m_j(\xi) \hat{f}(\xi).$$ Since $\sum_j m_j(\xi)$ converges to $m(\xi)$ pointwise a.e., it suffices to show that $\sum_j m_j(\xi) \hat{f}(\xi)$ is bounded pointwise by some $L^2$ function. Now, by the decay assumption on $m$, $m \in L^\infty$, so $m(\xi) \hat{f} \in L^2$. Since the $m_j$ are nonnegative and sum to $m$ pointwise, it follows that $$\sum_{j = -N}^N m_j \hat{f} \leq m\hat{f}\qquad \text{a.e.}$$ so since $m\hat{f} \in L^2$, we can conclude.