$$\frac{d}{d(2\cot\theta)}(\tan^{-1}2\cot\theta)\cdot\frac{d}{d(\cot\theta)}(2\cot\theta)\cdot\frac{d}{d\theta}(\cot\theta)\cdot\frac{d}{dx}\theta\tag{1}$$
$$\left(\frac{d}{d(2\cot\theta)}(\tan^{-1}2\cot\theta)\right)\cdot\left(\frac{d}{d(\cot\theta)}(2\cot\theta)\right)\cdot\left(\frac{d}{d\theta}(\cot\theta)\right)\cdot\frac{d}{dx}\theta\tag{2}$$
$$4\sin2x\cos2x\tag{3}$$
$$4(\sin2x)(\cos2x)\tag{4}$$
$(2)$ is completely unambiguous. No one can misinterpret anything. However, in $(1)$, one can misinterpret, for example, $\frac{d}{d(2\cot\theta)}(\tan^{-1}2\cot\theta)\cdot\frac{d}{d(\cot\theta)}(2\cot\theta)$ as meaning $\frac{d}{d(2\cot\theta)}\left[(\tan^{-1}2\cot\theta)\cdot\frac{d}{d(\cot\theta)}(2\cot\theta)\right]$ instead of $\left(\frac{d}{d(2\cot\theta)}(\tan^{-1}2\cot\theta)\right)\cdot\left(\frac{d}{d(\cot\theta)}(2\cot\theta)\right)$.
Similarly, $(4)$ is completely unambiguous. No one can misinterpret anything. However, in $(3)$, one can misinterpret $\sin2x\cos2x$ as meaning $\sin(2x\cos2x)$ instead of $(\sin2x)(\cos2x)$.
Is this a minor thing? Should I be worried about putting brackets obsessively? I experience so much anxiety now while doing math.
Just as spaces are important in separating words, so they are in mathematical notation. Thus $\sin2x$ means $\sin(2x)$, and $\sin2x\cos2x$ means $(\sin2x)(\cos2x)$. If we really mean $\sin(2x\cos2x)$, then we have to add the parentheses to override the spacing convention. To override the usual convention that multiplication (denoted by juxtaposition) operates before other operations, we can use a centered dot. Thus $\sin2\cdot x$ means $(\sin2)x$ rather than $\sin2x$.
The spacing convention implies that the notation $\tan^{-1}2\cot\theta$ means $(\tan^{-1}2)\cot\theta$. If we intend $\tan^{-1}(2\cot\theta)$, then the parentheses are necessary.
The dots in your expression #1 make it unambiguous, and there is no need to add the (big) parentheses as in expression #2. Also, the parentheses in $\dfrac{\mathrm d}{\mathrm d\theta}(\cot\theta)$ are redundant, but they are not redundant in $\dfrac{\mathrm d}{\mathrm d\theta}(2\cot\theta)$. Thus $\dfrac{\mathrm d}{\mathrm d\theta}\cot\theta=-\csc^2\theta$, while $\dfrac{\mathrm d}{\mathrm d\theta}2\cot\theta=\dfrac{\mathrm d}{\mathrm d\theta}2\cdot\cot\theta=0$.