Sectional curvature of product metric?

Yes, there is something that can be said.

Denote the connection on $(M, g_{1})$ by $\tilde{\nabla}$ and let $X_{1}, \ldots , X_{m}$ be an orthonormal frame on an open set $U \subset M$. Likewise, denote the connection on $(N, g_{2})$ by $\hat{\nabla}$ and let $Y_{1}, \ldots, Y_{n}$ denote an orthonormal frame on an open set $V \subset N$.

It follows that the connection $\nabla$ on the product manifold $M \times N$ satisfies

1. $\nabla_{X_{i}}X_{j} = \tilde{\nabla}_{X_{i}}{X_{j}}$
2. $\nabla_{X_{i}}Y_{j} = \nabla_{Y_{j}}X_{i} = 0$
3. $\nabla_{Y_{i}}Y_{j} = \hat{\nabla}_{Y_{i}}Y_{j}$.

Note that in the above that we are canonically identifying vector fields on the factors $M$ and $N$ with vector fields on the product $M \times N$.

Letting $Z_{i} = X_{i}$, $1 \le i \le m$, and $Z_{i + m} = Y_{i}$, $1 \le i \le n$, denote a corresponding orthonormal frame for $M \times N$, it follows immediately that the Riemannian curvature operator $R = R_{ijk}^{l}Z_{l}$ on $M \times N$ satisfies $R_{ijk}^{l} = 0$ unless $\left\{i, j, k\right\} \subset \left\{1, 2, \ldots, m\right\}$ or $\left\{i, j, k\right\} \subset \left\{m +1, m + 2, \ldots, m + n\right\}$.

It now follows almost immediately that $M \times N$ will have non-negative (or non-positive) sectional curvatures whenever both $M$ and $N$ do.

Note, however, that if $M$ and $N$ both have with everywhere positive sectional curvatures with respect to the metrics $g_1$ and $g_2$, then the product manifold $M \times N$ (equipped with the product metric) will have tangent two planes that have sectional curvature of zero. For example, consider the product manifold $S^{2} \times S^{2}$ where both factors of $S^{2}$ have the standard round metric that comes from their natural embeddings in Euclidean space.