i would like to show that the tangent bundle of $\mathbb{S}^{1}$ is diffeomorphic to a infinite cylinder $\mathbb{S}^{1}\times \mathbb{R}$. Well, the set of points $(p,v)\in T\mathbb{S}^{1}$ satisfies two equations: (i) $p_{1}^{2}+p_{2}^{2}=1$ and (ii) $p_{1}v_{1}+p_{2}v_{2}=1$, resolving this system in function of $(v_{1},v_{2})$ using the Cramer rule, i found that $v_1=-\lambda p_{2}$ and $v_{2}=\lambda p_{1}$, where $\lambda=\det \begin{bmatrix} q_{1} & q_{2} \\ v_{1} & v_{2} \end{bmatrix}$. Thus my candidate for diffeomorphism is $\phi: \mathbb{S}^{1}\times \mathbb{R} \to T\mathbb{S}^{1}$ such that $\phi(p,v)=(p,\lambda p^{t}$), where $p^{t}=(-p_{2},p_{1})$. I showed that $\phi$ is injective, smooth, but i don't know if this application is onto. Any tips?
My thinking is correct, how could I generalize it?
Your argument is correct, here is a different proof which generalize more easily : a vector bundle $E \to M$ of rank $r$ is trivial if and only if there are sections $s_1, \dots, s_n : M \to E$ such that for all $m \in M$, $(s_1(m), \dots, s_r(m))$ is a basis of $E_m$.
In particular, since the tangent bundle of $S^1$ is of rank $1$, this is enough to find a non-vanishing tangent vector field, which you did find, given by $(x,y) \mapsto (-y,x)$. Same vector field works for $T^n = S^1 \times \dots \times S^1$ and gives you $T((S^1)^n) = (S^1)^n \times \mathbb R^n$.
Notice that by Poincaré-Hopf theorem, any manifold with non-zero Euler characteristic (for example $S^2$ since $\chi(S^2) = 2$) can't have $TM \cong M \times \mathbb R^n$ ($n = \dim M$) since any vector field will have at least a zero somewhere.