You are missing a bit more.
Let $F: M\times {\mathbb R}\to M$ be the (partially defined) flow of the vector field $V$. By compactness of $S$, there exists $\epsilon>0$ such that you have a well-defined map
$$
\Phi: S\times (-\epsilon,\epsilon)\to M, \Phi(x,t)=F(x,t), x\in S.
$$
As you correctly noted, the map $\Phi$ is an immersion at each point of $S\times \{0\}$. Hence, taking $\epsilon>0$ small enough, you get that $\Phi$ is an immersion. [Indeed, immersion means injectivity of the derivative. But if the derivative is injective at one point, then it is also injective at all nearby points. Compactness of $S$ then implies the existence of $\epsilon>0$ such that $\Phi$ defined above is an immersion.]
Thus, for each $x\in S$ there exists $\epsilon_x\in (0,\epsilon)$ an a neighborhood $U_x$ of $x$ in $S$ such that the restriction
$$
\Phi: U_x\times (-\epsilon_x, \epsilon_x)\to M
$$
is 1-1. Since $\Phi$ restrict to the (identity) embedding on $S\times \{0\}$, for each pair $(x,y)\in S\times S$ there exists $\epsilon_{xy}\in (0,\epsilon)$ and a product neighborhood $U_{x,y}\times U_{y,x}$ of $(x,y)$ in $S\times S$ such that the restriction
$$
\Phi: (U_{x,y}\cup U_{y,x})\times (-\epsilon_{xy}, \epsilon_{xy})\to M
$$
is 1-1. [Indeed, the fact that $\Phi$ is an immersion implies that for each $x\in S$ there exists a product neighborhood $U_x\times (-\epsilon_x, \epsilon_y)$ of $(x,0)$ such that $\Phi$ restricted to this neighborhood is an embedding. Now, if $x=y$ then we can take $U_{x,y}= U_{y,x}=U_x$ and $\epsilon_{xy}=\epsilon_x$. Suppose that $x\ne y$. Then, after shrinking $U_x, U_y$ if necessary, we can find disjoint neighborhoods $V_x, V_y$ of $x, y$ in $M$ and $\epsilon_{xy}>0$ such that
$$
\Phi(U_x\times (-\epsilon_{xy}, \epsilon_{xy}))\subset V_x,$$
$$
\Phi(U_y\times (-\epsilon_{xy}, \epsilon_{xy}))\subset V_y.$$
This is a direct consequence of the fact that $\Phi$ restricted to $S$ is the identity map. Lastly, injectivity of $\Phi$ on each $U_x\times (-\epsilon_{xy},\epsilon_{xy})$, $U_y\times (-\epsilon_{xy}, \epsilon_{xy})$ implies the claimed injectivity of $\Phi$ on the union of these two open subsets. Hence, we can take
$$
U_{xy}=U_x, U_{y,x}=U_y,
$$
thereby proving the claim.]
By compactness of $S$, we can find a finite subcover of $S\times S$ by these product neighborhoods. Hence, there exists $\delta\in (0,\epsilon)$ such that
$$
\Phi: S\times (-\delta,\delta)\to M
$$
is 1-1. Since $S\times [-\delta/2,\delta/2]$ is compact and $M$ is Hausdorff, the injective continuous map
$$
\Phi: S\times [-\delta/2,\delta/2]\to M
$$
is a homeomorphism to its image. Thus, the restriction
$$
\Phi: S\times (-\delta/2,\delta/2)\to M
$$
is an immersion and a homeomorphism to its image, hence, an (smooth) embedding. qed
Best Answer
Solve for $x,y$ the equations $$\dot x=x\\ \dot y=-y$$ and so the integral curve would be $$\alpha(t)= (x(t),y(t))$$