As the comments indicate, the level of generality of the question is optimistic. Here's a special case: the product $M\times N$ of smooth manifolds.
Recall that a manifold $X$ is stably parallelizable if $TX\oplus \mathbb{R}$ is trivial. By a standard argument in obstruction theory, this is so as soon as $TX\oplus \mathbb{R}^n$ is trivial for some $n\geq 1$.
I assume $M$ and $N$ connected, positive-dimensional but not necessarily compact. The product $M\times N$ is parallelizable iff $M$ and $N$ are stably parallelizable and one of them has vanishing Euler characteristic.
Euler characteristics $\chi$ are relevant because $\chi(M\times N)=\chi(M)\chi(N)$ and because $\chi$ is precisely the obstruction to having one nowhere-zero vector field.
The "if" direction has a short, elementary proof that I'll leave to you, but you can also look it up in the extremely short paper of E. B. Staples, Proc. A.M.S. 18 no. 3 (1967). Conversely, if $M\times N$ is parallelizable then, choosing a trivialisation of $T(M\times N)$, and a point $y\in N$, we get by restriction to $M\times y$ a trivialization of $TM\oplus T_y N$. Hence $M$ (and similarly $N$) is stably parallelizable.
Part of this goes over to smooth fibre bundles $E \to B$ with connected base $B$ and fibre $F$: if $TE$ is trivial then $0=\chi(E)=\chi(B)\chi(F)$, and $F$ is stably parallelizable. But the pullback of the Hopf fibration $S^5\to \mathbb{CP}^2$ to $\mathbb{CP}^2\times S^1$ is an example where the total space $S^1\times S^5$ is parallelizable but the base is not stably parallelizable (it has non-vanishing $p_1$).