$\def\tr{\mathrm{tr}\;}\def\GL{\mathrm{GL}}\def\CC{\mathbb C}$ Let $\rho:G\to\GL_n(\CC)$ and $\sigma:H\to\GL_m(\CC)$ be two group homomorphisms. Let us name the components of the matrices in the image of $\rho$ in the obvious way so that we may write $\rho(g)=(\rho_{i,j}(g))_{1\leq i,j\leq n}$, and similarly for $\sigma$.
On the other hand, let $I=\{(i,j):1\leq i\leq n, 1\leq j\leq m\}$ and let us index the coefficients of a matrix $A\in\GL_{nm}(\CC)$ using $I$, so that we may write $A=(a_{(i,j),(k,l)})_{(i,j),(k,l)\in I}$. This makes sense, of course, because the matrices in $\GL_{mn}(\CC)$ have $mn=|I|$ rows and columns.
Consider the function $\lambda:G\times H\to\GL_{mn}(\CC)$ such that $\lambda(g,h)=(\lambda_{(i,j),(k,l)}(g,h))_{(i,j),(k,l)\in I}$ with $\lambda_{(i,j),(k,l)}(g,h)=\rho_{i,k}(g)\sigma_{j,l}(h)$ for all $(i,j)$, $(k,l)\in I$.
With some work —which is best done in private— one can show that $\lambda$ is an homomorphism of groups. A little computation, then, shows that for each $(g,h)\in G\times H$ we have $\tr\lambda(g,h)=\tr\rho(g)\cdot\tr\sigma(h).$ This means that the character of $\lambda$ is has the relation you want with the characters of $\rho$ and $\sigma$.