Product of Riemannian manifolds?

Yes. Using the natural isomorphism $T(M \times N) \cong TM \times TN$, define the metric on $T(M \times N)$ as follows for each $(p,q) \in M \times N$.

$$g^{M\times N}_{(p,q)} \colon T_{(p,q)}(M \times N) \times T_{(p,q)}(M \times N) \to \mathbb{R},$$ $$((x_1,y_1),(x_2,y_2)) \mapsto g^M_p(x_1,x_2) + g^N_q(y_1,y_2).$$

Alternately, if you think of the metrics as vector bundle isomorphisms $g^M \colon TM \to T^* M$ and $g^N \colon TN \to T^* N$, then the metric on $M \times N$ is just the induced vector bundle isomorphism

$$g^M \oplus g^N \colon TM \oplus TN \to T^*M \oplus T^* N,$$ $$(x,y) \mapsto (g^M(x), g^N(y))$$

Here, $TM \oplus TN$ is just $TM \times TN$ as a set, and the base manifold of $TM \oplus TN$ is $M \times N$.