Given a continuous differentiable functio $F:\mathbb{R}^n\mapsto \mathbb{R}^m$ with $n>m$. Define $$ {\cal M}=\{x\in\mathbb{R}^n: F(x)=0\} $$ and let $x_0\in{\cal M}$ such that the Jacobian of function $F$ at $x_0$, say $JF(x_0),$ has full-rank $m$.
Suppose that $r\in\mathbb{R}^n$ for which $JF(x_0)r=0.$ I wish to construct a path on ${\cal M}$ through $x_0$ with direction $r$ that means finding a continuous differentiable $p:(-a,a)\mapsto {\cal M}$ with small $a>0$ such that $p(0)=x_0$ and $p'(0)=r$.
Note tha if we define $p$ by $p(t)=x_0+tr$ then $p(0)=x_0$ and $p'(0)=r$ but $p(t)\notin{\cal M}$ because $$ F(p(t))=F(x_0)+tJF(x_0)r+o(|t|)=o(|t|) $$ and we know $o|t|$ is not really zero. So, there should be a modification on $p$. Any advice or suggestion?
As you have observed, there is not going to be any simple minded formula. Instead, you can apply the Implicit Function Theorem, the conclusion of which gives you a parameterization $\phi : U \to \mathcal{M}$, $\phi(0) = x_0$, where $U \subset \mathbb{R}^{n-m}$ is open. Once you have that, defining paths in $\mathcal{M}$ becomes easy: assuming (as you must) that the vector $r$ is in $\text{image}(d\phi_0)$, take $v = (d\phi_0)^{-1}(r)$, take $q(t)=tv$, then take $p(t) = \phi(q(t))$.