Let me first pick up where you were left (rightfully) confused.
$W$Net $=$ $m\int v\frac {dv}{dx}$ $dx$
And then the answer says "the $dx$ cancels, leaving you with
$m\int vdv$
However, as I understand it, the $dx$ can't just "cancel" [...]
Yes, you are completely correct in saying that you can't just blindly say "it cancels" (as many people and textbooks might do). Since we're working with an integral, the correct way to proceed is to consider substitution of variables.
We have the integral
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v\frac{dv}{dx} \, dx $$
where I decided to include the definite limits of integration. Here $x_{i}$ is the initial position and $x_{f}$ is the final position.
Evidently the variable $v$ is meant to be taken as a function of $x$ in the integrand, so we'll write it out as
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v(x)v'(x) \, dx. $$
Now look at the wikipedia article for how substitution is properly done: https://en.wikipedia.org/wiki/Integration_by_substitution#Definite_integrals
$$ \int_a^b f(\varphi(x))\varphi'(x)\, dx = \int_{\varphi(a)}^{\varphi(b)} f(u)\,du. $$
We can match our integral to the LHS integral by taking $f(w) = w$ (so $f$ is just the identity function) and $\varphi(x) = v(x)$. Then
$$ W_{\text{Net}} = m\int_{x_{i}}^{x_{f}} v(x)v'(x) \, dx = m\int_{v(x_{i})}^{v(x_{f})} u \, du. $$
This last integral can be evaluated directly and I recommend you do it yourself. You should immediately obtain the work-energy theorem.
However, if you wish, we can do some "notation cleaning" to match what is written in physics texts. First, we know $v(x_{i})$ is the velocity at initial position of the path, so it is just the initial velocity — we'll label it as $v_{i} = v(x_{i})$. Similarly we label the final velocity as $v_{f} = v(x_{f})$. Inside the integrand, the letter being used doesn't matter, but since the limits of the integral are velocities, we know that the meaning of the variable being used is velocity (to be more specific $du$ must have units of velocity). Thus, it makes sense to relabel $u$ as the letter $v$ to carry the connotation of velocity. From this we get
$$ W_{\text{Net}} = m\int_{v_{i}}^{v_{f}} v \, dv = m \left( \frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} \right) = \frac{1}{2}mv_{f}^{2} - \frac{1}{2}mv_{i}^{2}, $$
which is the work-energy theorem, as desired.
Now let me address another point that you raised.
[...] and furthermore, shouldn't the integral of
$\int v\frac {dv}{dx}$ $dx$ when integrating over $dx$ be equal to $(\frac{v^2}{2})(v)$, since the integral of $\frac{dv}{dx}$ is just $v$ and the integral of $v$ should be $\frac{v^2}{2}$?
No, this is wrong. When you have a product inside an integral, you cannot integrate each factor individually. Explicitly,
$$ \int_{a}^{b} f(x)g(f) dx \ne F(x)G(x)|_{a}^{b}. $$
Instead, the usual right way to handle products is to use integration by parts. Now it is not always needed, and in our case we definitely don't need it.
What I hope you understand is that this talk of "canceling $dx$ in the integral" is a actually shorthand for u-substitution just like how "treating derivatives like fractions" is a shorthand for the chain rule.
As you can see, it's a bit more complicated when you do all this rigorously.
I applaud your insistence on rigor here, and I hope you continue to probe these details. It will further your understanding. However, at the same time I should also say that there is value in these shorthands, and you should know how to work with them as well. Otherwise, you will end up far too inflexible.
True understanding comes from being comfortable with both approaches (the rigorous way and the handwavy shorthand way) and knowing how to synthesize them together.
