Let $(M,h)$ be a Riemannian space and given some $x_0\in M$, consider the geometric Riemannian sphere of radius $r$ centered at $x_0$, i.e.
$$S^r_h(x_0)=\{x\in M| d_h(x_0,x)=r\}.$$
Let $W_0$ be a constant vector tangent to $M$ s.t. $||W_0||<1$. Let $S=S^r_h(x_0)+\varphi(r,x_0)$ and note that $x\in S^r_h(x_0)$ if and only if $x+\varphi(r,x_0)\in S$. Here $\varphi(s,x)$ is the flow of $W_0$, i.e., $\frac{d\varphi}{ds}=W_0(\varphi)$.
One can say that $$T_{x+\varphi(r,x_0)}S=T_xS^r_h(x_0)+rW_0,$$ i.e. $T_{x+\varphi(r,x_0)}S$ is parallel to $T_xS^r_h(x_0)$?
Practically it seems affirmative however I could not prove it. Any comment?
By $T_xS^r_h(x_0)$ I mean the tangent space of $x$ and so on.
I am wondering what I am doing is technically correct?
You still have some problems in your post, but now it is clear to me that the problem is that you are being misled by your $\Bbb R^n$ intuition, and not just using a notational shorthand. This can indeed be stated in terms of flows, but not as you currently have it.
You describe $W_0$ as a "constant vector tangent to $M$". Since you now speak of its flows, apparently you are considering $W_0$ to be defined everywhere in $M$ (or at least on some neighborhood). But there is no such thing as a "constant" vector field. The tangent spaces $T_xM$ and $TyM$ for distinct points $x, y \in M$, are entirely separate vector spaces, with no intrinsic way of relating vectors in $T_xM$ to vectors in $T_yM$. What does "constant" mean when there is no relationship?
The most obvious way to fix it is simply to drop the word "constant", and take $W_0$ to be an arbitrary smooth vector field on some neighborhood of $x_0 \in M$ containing $S_h^r(x_0)$ satisfying $W_0 \ne 0$ everywhere (that $\|W_0\| < 1$ is not needed). That $W_0 \ne 0$ is crucial, as the flow $\varphi$ will be singular where $W_0 = 0$. But this is also why we can't just have $W_0$ defined on all $M$. There may not be a never-zero vector field on $M$. Quite famously, you can't comb the hair on a sphere.
A second issue is that you are still thinking of $M$ itself as having an addition defined on it. "$S=S^r_h(x_0)+\varphi(r,x_0)$" and "$x+\varphi(r,x_0)\in S$" are non-sensical. There is no definition of "$+$" between the elements of $M$.
Fortunately, in this case what you are after is clear. $S = \varphi(S^r_h(x_0), r)$, the image of $S^r_h(x_0)$ under the mapping $\varphi_r : x \mapsto \varphi(r,x)$. (Note that the domain of $W_0$ must contain all of $S^r_h(x_0)$
Now we get to the result you want to show: $$T_{x+\varphi(r,x_0)}S=T_xS^r_h(x_0)+rW_0$$ but again, it doesn't make sense, even after fixing the basepoint addition: $T_{\varphi_r(x)}S$ and $T_xS_h^r(x_0)$ are separate vector spaces. You cannot add anything to vectors in $T_xS_h^r(x_0)$ to get a vector in $T_xS_h^r(x_0)$, because no such addition is defined. Now $S = \varphi_r(S^r_h(x_0))$, and you can show that because $W_0 \ne 0$ anywhere, $\varphi_r$ is not singular. Therefore $$T_{\varphi_r(x)}S = d\varphi_r(T_xS^r_h(x_0))$$