The cardinal product of two sets is defined to be the cardinality of the Cartesian product. The Cartesian product is:
$$\prod_{\alpha \lt\beta}\kappa_{\alpha}=\{f\mid f\colon\beta\rightarrow (\cup_{\alpha\lt\beta}A_\alpha)\land(\forall_{\alpha}\lt\beta)(f(\alpha)\in A_{\alpha})\}$$
where $\langle\kappa_{\alpha}\mid\alpha < \beta\rangle$ is a sequence of sets.
It is said that there exists a bijection between this and the usual notion of a Cartesian product. But, I can't see how. If $X$ and $Y$ are two sets, where the former contains five elements and the latter contains three, then the definition stated would create a function with two elements. It would look something like:
$\{(1,x), (2, y)\}$, where $x\in X$ and $y\in Y$
The usual notion of a Cartesian product would have fifteen. How can there be a bijection if the amount of elements in each set is different?
If this is the case, then the cardinal product, 5 x 3, could either be 15 or 2. Which one is it? Is it neither? What am I not understanding?
Note that $X_0\times X_1=\{(x_0,x_1)\mid x_i\in X_i\}$. The product $\prod_{i<2}X_i=\{f\colon 2\to\bigcup\{X_0,X_1\}\mid f(i)\in X_i\}$, it is not a function, it is a set of functions.
There is a natural bijection between the two sets. Namely, $(x_0,x_1)\mapsto\{(0,x_0),(1,x_1)\}$. You seem to have confused a typical element of $\prod_{i<2}X_i$, which is indeed a set with two elements, with the entire product.
The mapping is injective, since if $(x_0,x_1)\neq(x_0',x_1')$ then either $x_0\neq x_0'$ or $x_1\neq x_1'$ and in that case the resulting function is different; and the mapping is surjective since if $\{(0,x_0),(1,x_1)\}$ is a function in the product, then the pair $(x_0,x_1)$ is mapped to it.
In particular, if $X_0$ has five elements, and $X_1$ has three elements, $\prod_{i<2}X_i$ has $15$ elements.