In the derivation of the formula for surface integrals we find that the surface $S$ of a parametric function $f(u,v)$ for the area $D$ can be found using the following entity:
$$A(S) = \iint_{D} \left \| \frac{\partial f(u,v)}{\partial u} \times \frac{\partial f(u,v)}{\partial v}\right \| du dv$$
The derivation of this formula is crystal clear to me, so I do not need any explanation on that topic.
The formula for the length of a cross product of vectors (and hence the surface of the parallelogram determined by $\vec{u}$ and $\vec{v}$) is:
$$\left \| \vec{u} \times \vec{v} \right \| = \left \| \vec{u} \right \|\cdot \left \| \vec{v} \right \|\cdot \sin \theta $$
As soon as the integral is being worked out in my coursework (and elsewhere), the $\sin \theta $ part however is magically dropped. This leaves me to only one conclusion: $$\sin \theta = 1 \Rightarrow \theta = \frac{\pi}{2}$$
If this were true, this would just mean that the parallelograms we used to define a surface integral just became squares, whilst no explanation is ever given (I tried to search on this topic online, but my treasure hunt was fruitless). Is there any reason and proof why we can just drop out the $\sin \theta $ part?
I suspect this could be true because of this integral being an infinite sum of infinitesimal small surfaces, which can be reasoned as squares, but I seem to fail to find a mathematical proof for that assumption and thus it stays just a wild assumption.
This is not true. Note that $\theta = \frac{\pi}{2}$ is equivalent to say that for all $u,v$ $$\frac{\partial f}{\partial u}(u,v) \cdot \frac{\partial f}{\partial v}(u,v) = 0$$ where $\cdot$ denotes the scalar product of vectors.
Now, a counterexample follows: let $f(u,v) = (v\cos^2 u , v\sin u , v)$ with $(u,v) \in (a,b) \times (c,d)$ (actually the domain of $f$ is not important, since your question is concerning about a local fact).
Then $$\frac{\partial f}{\partial u}(u,v) \cdot \frac{\partial f}{\partial v}(u,v) = v \cos u \sin u (1 - 2 \cos^2 u) \neq 0$$ For example, take $u=1, v=1$, you can see that this does not vanish.