Here are some suggestions that may help.
$\bullet$ How to self study Linear Algebra. See my response and the other responses.
Study Habits from Math Stack Exchange
$\bullet$ What are some good math study habits?
$\bullet$ How to study math to really understand it and have a healthy lifestyle with free time?
$\bullet$ How can I learn to read maths at a University level?
Free Resources
$\bullet$ MIT Open Course Ware. You can also find others at other universities and these have free course notes and video lectures.
$\bullet$ Khan Academy.
$\bullet$ Wiki Books Linear Algebra Resources
$\bullet$ Linear Algebra for Communications: A gentle introduction. I would study this to give you context!
$\bullet$ College Libraries: Peruse the books in the library and see if any fit the style you like.
Book Recommendations
$\bullet$ Linear Algebra Through Geometry, Thomas Banchoff, John Wermer (since you are visual)
$\bullet$ Practical Linear Algebra: A Geometry Toolbox, Gerald Farin, Dianne Hansford (since you are visual)
$\bullet$ Linear Algebra Done Right, Sheldon Axler
$\bullet$ Linear Algebra and Its Applications, David C. Lay
$\bullet$ Linear Algebra Textbook Recommendations on MSE
$\bullet$ Alternative to Axler's “Linear Algebra Done Right”
Problem Books
It is very helpful to do problems and practice, practice, practice to reinforce concepts!
$\bullet$ Linear Algebra Problem Book, Paul R. Halmos
$\bullet$ The Linear Algebra a Beginning Graduate Student Ought to Know, Jonathan Golan.
$\bullet$ 3,000 Solved Problems in Linear Algebra, Seymour Lipschutz
$\bullet$ Schaum's Outline of Linear Algebra Fourth Edition, Seymour Lipschutz, Marc Lipson
History of Linear Algebra
Linear Algebra History Websites
Regards
It's true that one cannot assume that roots are multiplicative over the complex numbers. But I don't think that's an issue here.
You don't show your second computation. I have
$$
-\operatorname{cis}(\tfrac\pi2)=\operatorname{cis}(\pi)\,\operatorname{cis}(\tfrac\pi2).
$$
If you write the cubic roots the usual way, you get
$$
\operatorname{cis}(\tfrac\pi3+\tfrac{2k\pi}3)\operatorname{cis}(\tfrac\pi6+\tfrac{2\ell\pi}3)=\operatorname{cis}(\tfrac\pi3+\tfrac\pi6+\tfrac{2(k+\ell)\pi}3)=\operatorname{cis}(\tfrac\pi2+\tfrac{2(k+\ell)\pi}3)
$$
Best Answer
This is pretty soft, but I saw an analogy online to explain this once.
If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true. $$\text{backward} \times \text{backward} = \text{forward}$$ $$ \text{negative} \times \text{negative} = \text{positive}$$ It's not perfect, but it introduces the notion of the number line having directions at least.