Let $(X, d)$ be a metric space and suppose $k: X\times X \to \mathbb{R}$.
A paper I am reading says that "$k(\cdot, \cdot)$ is a continuous function". However, I am not familiar with continuity of multivariate functions on products of metric spaces. Does this mean that $k$ is continuous with respect to the product metric
$$ d^*((x_0,x_1), (x'_0, x'_1)) = \sqrt{d(x_0,x'_0)^2 + d(x_1, x'_1)^2}, $$
so that
$$ \forall \epsilon > 0,\; \exists \delta > 0,\; d^*((x_0,x_1), (x'_0, x'_1)) < \delta \implies |k(x_0,x_1) - k(x'_0,x'_1)| < \epsilon\,? $$
If this interpretation is correct, is there a canonical textbook reference that anyone can recommend which clarifies this particular construction?
The main point is that the product of metric spaces is a metric space (like the product of vector spaces is a vector space, and the product of groups is a group, etc). The definition of "product of metric spaces" must specify how the Cartesian product of the underlying sets is equipped with a metric. A typical choice is the product metric you mentioned, e.g., Prove that the product space is a metric space.
It's true that there are other natural choices such as $$d_1((x_0,x_1), (x'_0, x'_1)) = d(x_0,x'_0) + d(x_1, x'_1)$$ and $$d_\infty((x_0,x_1), (x'_0, x'_1)) = \max( d(x_0,x'_0), d(x_1, x'_1))$$ It's easy to see that the metrics they define on the product are comparable to $d^*$: $$ d_\infty \le d^* \le d_1 \le 2d_\infty \tag1 $$
In particular, they induce the same topology but (1) is much stronger than that: this kind of equivalence (bi-Lipschitz equivalence) preserves most of the metric structure beyond topology (completeness, Lipschitz functions, Hausdorff dimension, etc).
The point being, the metric structure of product space is a thing to establish first. Then there is no need to discuss "continuity of multivariate functions on products of metric spaces"; the continuity is just regular continuity on the metric space $X\times Y$.
Since you asked for references, here is one: "A Course in Metric Geometry" by Burago, Burago, Ivanov.