Surface integral, why do we consider only the normal component?

Whenever we integrate a vector field over a suface, we consider an elemental area and we dot product the area with the vector field equation and then integrate it.But by this method we are adding up the component of the field perpendicular to the area at every point on the surface(normal).But what about the other component? Why does the other component does not contribute to the effect? If it is velocity field, does the particle break into two components and one part goes normally into the suface and the otger goes tangentially out the surface?

In the time interval $dt$, the particles with velocity tangent to the surface will "remain" on the surface, and do not contribute to the outward flux, computed from $t$ to $t+dt$.
Their outward displacement will be of order $dt^2$.

Those particles are considered to "remain" within the considered volume, subject to the interaction with all the other still "inside" at time $t+dt$.

Your surface $S$ is covered by two families of parameter lines that form a web of tiny paralelograms. Let's look at such a parallelgram with a microscope.

We see a homogeneous (i.e., spatially constant) flow field ${\bf v}$ and a wire frame $F$ in the form of a parallelogram spanned by two vectors ${\bf a}$ and ${\bf b}$. The area enclosed by this frame is given by $${\rm area}(F)=|{\bf a}\times{\bf b}|\ ,$$ and the unit normal ${\bf n}$ of $F$ is given by $${\bf n}={{\bf a}\times{\bf b}\over |{\bf a}\times{\bf b}|}\ .$$ If the "fluid" is actually toothpaste the amount of substance flowing through the frame per second fills the "totally skew" parallelepiped $P$ spanned by ${\bf a}$, ${\bf b}$ and ${\bf v}$. Note that $P$ is not "standing upright" on $F$. If, e.g., ${\bf v}$ is almost parallel to $F$ then ${\rm vol}(P)$ will be small indeed. One has $${\rm vol}(P)=[{\bf a},{\bf b},{\bf v}]=({\bf a}\times{\bf b})\cdot{\bf v}={\rm area}(F)\>{\bf n}\cdot{\bf v}\ .\tag{1}$$ whereby one has to take proper care of the orientation of ${\bf n}$. The formula $(1)$ can be interpreted as "toy model" of a flux integral: The flux $\Phi$ of ${\bf v}$ through $F$ is given by $$\Phi:=\int_F{\bf v}\cdot{\bf n}\>{\rm d}\omega\ ,$$ whereby ${\rm d}\omega$ denotes the scalar surface element. Note that the actual computation of $\Phi$ does not involve square roots, only a triple vector product, see $(1)$.