If a smooth $n$-manifold $M \subset \mathbb{R}^m$ is globally defined as the preimage of a regular value of a smooth function $f:\mathbb{R}^n \to \mathbb{R}^{m-n}$, then we have the desired setup
$$TM=\{(x,v) \in \mathbb{R}^{m} \times \mathbb{R}^{m} \mid f(x)=0 \text{ and } df_{x}(v)=0\}.$$
If you read his answer more carefully, @jef808 was saying that we could try to use the defining equations to find linearly independent global sections. For example, one obvious solution to the equation $ux+vy=0$ is $(u,v)=(-y,x)$. Then the map $S^1\to TS^1$ given by $(x,y) \mapsto ((x,y),(-y,x))$ defines a nonvanishing vector field. We can generalize this construction to $S^{2n-1}$ and its tangent bundle
\begin{align*}
TS^{2n-1} &= \bigg \{((x_1,y_1,\ldots,x_n,y_n),(u_1,v_1,\ldots,u_n,v_n)) \in \mathbb{R}^{2n} \times \mathbb{R}^{2n} \mid \\
& \qquad \qquad \sum_{k=1}^n x_k^2 +\sum_{k=1}^n y_k^2=1 \text{ and } \sum_{k=1}^n u_k x_k + \sum_{k=1}^n v_k y_k =0 \bigg \}
\end{align*}
by defining a map
$$(x_1,y_1,\ldots,x_n,y_n)\mapsto ((x_1,y_1,\ldots,x_n,y_n),(-y_1,x_1,\ldots,-y_n,x_n)).$$
However, for $n >1$, we need several linearly independent nonvanishing sections. As we found out circa 1960, this is only possible for $S^1$, $S^3$, and $S^7$; try googling "parallelizability of spheres". So even when the defining equations are very simple, it may be practically impossible to see whether or not you can construct enough linearly independent global sections of the tangent bundle.
Similarly, if there was another way to use the defining equations to demonstrate triviality of a bundle, then Milnor, Bott, et cetera probably would have used it in the case of spheres.
One can ask if there are special cases where the defining equations shed light on the issue. I don't know if there are interesting families of special cases, but there are certainly ad hoc methods on a case-by-case basis:
Example (Torus). Even though we know that the torus has a trivial tangent bundle, it's not so apparent from a particular embedding $M \subset \mathbb{R}^3$, say
$$M=\{(x,y,z) \in \mathbb{R}^3 \mid f(x,y,z)=(x^2 +y^2+z^2+3)^2-16(x^2+y^2)=0\}.$$
We know that points $((x,y,z),(u,v,w))$ in the tangent space satisfy
\begin{align*}
df_{(x,y,z)}(u,v,w)&=\begin{bmatrix} -32x +4x(3+x^2+y^2+z^2) \\ -32y +4y(3+x^2+y^2+z^2) \\ 4z(3+x^2+y^2+z^2)
\end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} \\
&=-16(2x u+2yv)+2(3+x^2+y^2+z^2)(2xu+2yv+2zw) \\
&=0.\end{align*}
Similar to the $S^1$ case, we have an obvious solution of $(u,v,w)=(-y,x,0)$. Call this first vector field $\nu$. To find a second (linearly independent) vector field $\eta$, we could start by assuming $\eta=(u,v,w)$ is perpendicular to $\nu=(-y,x,0)$, i.e. $\eta=c_1(x,y,0)+c_2(0,0,1)$ for some $c_1,c_2$ (that are functions of $x,y,z$). Plugging $u=c_1 x$, $v=c_1 y$, $w=c_2$ into the above equation, we must have
$$-16(2x^2 c_1+2y^2 c_1)+2(3+x^2+y^2+z^2)(2x^2 c_1+2y^2c_1+2z c_2) =0,$$
equivalently written as
\begin{align*}2x^2 c_1+2y^2c_1+2z c_2 &=\frac{16(2x^2 c_1+2y^2 c_1)}{2(3+x^2+y^2+z^2)} \\
c_2 &=\frac{c_1}{z} \left(\frac{8(x^2 +y^2 )}{3+x^2+y^2+z^2} -x^2 +y^2\right).
\end{align*}
If we set $c_1=z$, this gives us a reasonably simple expression for $c_2$. One can easily check that $\nu(x,y,z)=(-y,x,0)$ and $\eta(x,y,z)=(x z, y z, c_2(x,y,z))$ define two linearly independent nonvanishing vector fields on $M$. Thus $M$ has a trivial tangent bundle.
Best Answer
Here are three ways.