Parallelizable product of manifolds 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.
 A: 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 $f^*\xi \cong \eta$
