Indeed, $\cot^2(a)$ means
$$\left(\cot (a)\right)^2.$$
You need to apply the Chain Rule twice: first, to deal with the square: set $g(u)=u^2$ as your "outside function", and $u=f(\theta) = \cot(\sin(\theta))$ as your inside function. Since $g'(u) = 2u$, then
$$\frac{d}{d\theta}\cot^2(\sin(\theta)) = \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2 = g'(u)f'(\theta) = 2uf'(\theta) = 2\cot\bigl(\sin(\theta)\bigr)f'(\theta).$$
Now let's deal with $f'(\theta)$; we have $f(\theta) = \cot\bigl(\sin(\theta)\bigr)$. The "outside function" is $h(u) = \cot(u)$, the "inside function" is $u(\theta) = \sin(\theta)$. Since $h'(u) = -\csc^2(u)= -\left(\csc(u)\right)^2$, and $f'(\theta) = \cos(\theta)$, we have:
$$\frac{d}{d\theta}\cot\bigl(\sin(\theta)\bigr) = h'(u)u'(\theta) = -\csc^2(\sin\theta)\cos(\theta).$$
Putting it all together:
$$\begin{align*}
\frac{d}{d\theta}\cot^2\bigl(\sin(\theta)\bigr) &= \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2\\
&= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)\\
&= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \left(-\csc^2\left(\sin(\theta)\right)\left(\frac{d}{d\theta}\sin(\theta)\right)\right)\\
&= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot\left(-\csc^2\left(\sin(\theta)\right)\cos(\theta)\right)\\
&= -2\cot\bigl(\sin(\theta)\bigr)\csc^2\bigl(\sin(\theta)\bigr)\cos(\theta).
\end{align*}$$
I think I see the core of your question from your last paragraph, your image, and your response in post. The essential answer to your question boils down to the definitions of the trig functions.
The definitions of the trigs are nothing more than definitions, and they are defined over right triangles with $\theta$ being either of the non-right angles. That is to say for example, as you surely know, $\sin\theta=\frac{opposite}{hypoteneuse}$. By geometric principles, given any two sides of any right triangle we can always find the third side by the theorem of Pythagoras.
The technique of trigonometric substitution (like any other substitution) works when we can make a total substitution, and then completely reverse substitute the final result. In the case of succesful trig substitutions, we can always go back and call out the value of any of the trigs including $\cot \theta$. We can draw these because ALL of the trigs are defined over right triangles, and we can always find a third side given two. No one trig is more difficult to find than another over a right triangle with defined side lengths.
When all of the terms of an integral can be completely substituted by the method, the result, assuming you can find it, should always be re-translatable in to the original terms as all of the information required for the back substitution is included in the contents of a right triangle, as a trig of an angle is nothing more than the ratio of two sides of a right triangle.
Given any of the 6 trigs with variable theta, we can always assign measures to two of three sides of a right triangle as that is how the trigs are defined, hence we can always define the third side.
I hope this post is helpful, please let me know if it is less than clear. Trig sub is my favorite technique as far as methods go and so I am happy to have this out with you.
Best Answer
The chain rule does say something about integrals, but not what you seem to think. The chain rule says $$ \dfrac{d}{dx}f(g(x)) = f'(g(x)) g'(x)$$ Integrating both sides gives you $$ \int f'(g(x)) g'(x)\ dx = f(g(x)) + C$$ Now you can't just divide out the $g'(x)$ from the left side, because that $g'(x)$ is inside the integral: $\dfrac{1}{g'(x)} \int f'(g(x)) g'(x) \ dx$ is not the same as $\int f'(g(x))\ dx$.