Michael Hong, as always an excellent drawing, thank you!
Concerning
\$$\otimes((f \circ f'),(g \circ g')) = \otimes(f,g) \circ \otimes(f',g')\$$

1. We have a product \$$C \times C\$$, which has as objects ordered pairs \$$(x, y)\$$, and morpisms between such pairs, also ordered pairs like \$$(f, g)\$$, so \$$C \times C((x, y), (x', y')) = C(x, x') \times C(y, y'))\$$. This is all we know from definition.

2. Let's say the functor \$$\otimes\$$ maps an ordered pair \$$(f, g) \$$ to some morphism \$$a: C \to C\$$, and \$$(f', g')\$$ to \$$b: C \to C\$$, like this: \$$\otimes(f, g) = a\$$ and \$$\otimes(f', g') = b\$$, and of course preserves composition \$$\otimes((f' g') \circ (f, g)) = \otimes((f',g')) \circ \otimes((f,g))\$$.

What I need to figure out what is \$$(f' ,g') \circ (f, g)\$$. We have a lot of morphisms to choose from - \$$C \times C((x, y), (x'', y'')) = C(x, x'') \times C(y, y''))\$$, and among them, by definition of category, we have a morphism \$$(f' \circ f, g' \circ g)\$$. So we picked it up as the only candidate for \$$(f', g') \circ (f, g)\$$, which makes perfect sense, but still I'm not sure how to show this.