This problem comes from Hirsh's book Differential Topology, chapter 4, section 2, question 5.
Let $M$ and $N$ be (paracompact) $C^r$ manifolds, and $\epsilon^n_E$ the trivial vector bundle on $E$. If $TM$ admits a nonvanishing section, and $TM \oplus \epsilon^1_M$ and $TN \oplus \epsilon^1_N$ are trivial, then the manifold $M \times N$ is parallelizable.
I went for studying short exact sequences of vector bundles however I did not see how to use the section in this scenario and although I got morphisms from $N \times M \times \mathbb{R}^l$ to $T(M \times N)$ and vice versa but I never got any isomorphism.
For example, if $TN \oplus \epsilon^1_N \simeq N \times \mathbb{R}^n$ and $TM \oplus \epsilon^1_M \simeq M \times \mathbb{R}^m$ I can get an morphism $N \times M \times \mathbb{R}^{r+m} \rightarrow T(M \times N)$. Indeed by composing product maps and isomorphism I get the maps: $$\rho : N \times M \times \mathbb{R}^{r+m} \twoheadrightarrow M \times \mathbb{R}^m \simeq TM \oplus \epsilon^1_M \twoheadrightarrow TM $$ $$\phi : N \times M \times \mathbb{R}^{r+m} \twoheadrightarrow N \times \mathbb{R}^n \simeq TN \oplus \epsilon^1_N \twoheadrightarrow TN $$
Then the morphism is $(\rho, \phi)$. In the other way, we get another morphism $T(M \times N) \rightarrow N \times M \times \mathbb{R}^{r+m}$ through the inverse inclusion maps. I do not see however how to use the nonvanishing section as it does not guarantee a monomorphism $M\times \mathbb{R}^{dim(M)} \hookrightarrow TM$ since it is just nonvanishing, nor do I see how to get an isomorphism out of this commutative diagram. I really would appreciate a tip.
Thank you
Update :
I did found that the section gives an isomorphism $TM \simeq \epsilon^1_M \oplus \eta$ with $\eta$ a vector bundle of base $M$. By the first properties we get $\epsilon^2_M \oplus \eta \simeq M \times \mathbb{R}^n \simeq TM\oplus \epsilon_M^1$. Then in this case we get the sequence.
Then we get the right exact sequence : $$ \epsilon^2_M \oplus \eta \twoheadrightarrow \epsilon^1_M \oplus \eta \rightarrow 0 $$ By the definition of the quotient of vector bundles we get a unique bundle $\zeta$ such that $\frac{\epsilon^2_M \oplus \eta}{\zeta} \simeq \epsilon^1_M \oplus \eta \simeq TM$. This $\zeta$ is also the kernel of the morphism $\epsilon^2_M \oplus \eta \twoheadrightarrow \epsilon^1_M \oplus \eta $ as such it is isomorphic to $\epsilon^1_M$. Thus we get $$TM \simeq \frac{\epsilon^2_M \oplus \eta}{\epsilon^1_M} \simeq \frac{TM\oplus \epsilon_M^1}{\epsilon^1_M} \simeq \frac{M \times \mathbb{R}^m}{M \times \mathbb{R}}$$ From there nothing tells us that quotient of trivial bundles are trivial.
To answer the question I'll use the hint given by Jason De Vito in his comment. First let's show that : $$\pi^*_M(TM) \oplus \pi^*_N(TN) \cong T(M \times N)$$ Indeed $T(M \times N)$ is indeed a bundle, and the whitney sum can be expressed as a pull back bundle of the diagonal morphism between $M \times N$ and the vector bundle $(p, T(M)\times T(N), M \times N \times M \times N )$ however through the universal property of pullback bundles, since $T(M\times N) \rightarrow T(M) \times T(N)$ which extends the diagonal map provides an isomorphism between $\pi^*_M(TM) \oplus \pi^*_N(TN) \cong T(M \times N)$. In the same manner, we get a bundle morphism $(\pi_M^*TN) \oplus \epsilon^1_{M \times N}$ to $TN \oplus \epsilon^1_N$ by summing the projection $\epsilon^1_{M \times N} \twoheadrightarrow \epsilon^1_N$ and the morphism we got before $(\pi_N^*TN) \rightarrow TN$ That indeed covers the projection on $N$ gives us an isomorphim again by the universal property : $(\pi_N^*TN) \oplus \epsilon^1_{M \times N} \cong TN \oplus \epsilon^1_N$. By the exact same reasoning combined by the fact that $\eta \oplus \epsilon_{M \times N}^1 \cong TM$. We finally get :
$$\pi^*_M(TM) \oplus \pi^*_N(TN) \cong \pi^*_M(\eta) \oplus \epsilon_{M \times N}^1 \oplus \pi_N^*TN \cong \pi^*_M(\eta) \oplus \pi^*_N(TN \oplus \epsilon^1_N ) \cong \pi^*_M(\eta) \oplus \pi^*_N(\epsilon^n_N) \cong \pi^*_M(\eta) \oplus \epsilon^n_{N \times M} \cong_{\text{by induction}} \pi^*_M(\eta \oplus \epsilon^2_M) \oplus \epsilon^{n - 2}_{N \times M} \cong \epsilon^{m + n -2}_{M \times N} $$
Which completes the proof. ($\pi_N(\epsilon^n_N) \cong \epsilon^n_{N \times M}$ comes by taking $TN = \epsilon^1_N$)
The universal property of the bundle pullback I refer to is the following : If $f : M_0 \rightarrow M$ and $\xi$ a vector bundle over $M$ then for every vector bundle $\eta$ over $M_0$ that have a morphism over $f$ to $\xi$ there exist an isomorphism $f^*\xi \cong \eta$