I'm reading this subject as a hobby. Could someone help me solve this problem, since I have been doing some geometry for some time?
Let $\mathbf{e}_{3}$ be smooth unit normal along the immersion $\mathbf{x} \colon M \to \mathbb{R}^3$ compatible with the orientation of $M$.
If $g$ is a smooth function with compact support $S\subset M$ then there exist $\epsilon >0$ such that $$ X \colon M \times (-\epsilon, \epsilon) \to \mathbb{R}^3, \quad X(m,t) = \mathbf{x}(m) + t g(m) \mathbf{e}_3(m) $$ is an admissible variation of $\mathbf{x}$. If $a$ and $c$ are the principal curvatures of $\mathbf{x}$, then $\epsilon = \min_{overS}\left\{ \frac{1}{|a|},\frac{1}{|c|}\right\} $ works.
The definition of admissible variation is as follows:
Definition 8.1. An admissible variation of $\mathbf{x}$ is any smooth map $$ X \colon M \times (-\epsilon, \epsilon) \to \mathbb{R}^3, $$ with compact support, such that for each $t \in (-\epsilon, \epsilon)$, the map $$ \mathbf{x}_t \colon M \to \mathbb{R}^3, \quad \mathbf{x}_t(m) = X(m,t), $$ is an immersion. The support of $X$ is the closure in $M$ of the set of points of $M$ where $\mathbf{x}_t(m) \neq \mathbf{x}(m)$, for some $t$.
(Original scanned image here.)
This would be correct if you had no function $g$ in the variation, i.e., if $g=1$ everywhere. You'll need to divide your $\epsilon$ by the maximum of $|g|$ on $M$. Let's call that quantity $C$.
Then the result is easy enough to prove. To simplify things, assume there are no umbilic points and let $\mathbf e_1,\mathbf e_2$ be a principal moving frame, with dual coframe $\omega_1,\omega_2$. Then $d\mathbf e_3 = -(k_1\omega_1\mathbf e_1 + k_2\omega_2\mathbf e_2)$, and \begin{align*} d\mathbf x_t &= d\mathbf x + tg d\mathbf e_3 + t\,dg\,\mathbf e_3 \\ &= (1-tgk_1)\omega_1\,\mathbf e_1 + (1-tgk_2)\omega_2\,\mathbf e_2 + t\,dg\,\mathbf e_3. \end{align*} This will have rank $2$ (independent of the nature of the function $g$) provided $|t|<\dfrac 1{C\sup|k_i|}$ for $i=1,2$.