Bounds on derivative of a kind of distance function Suppose we have some closed manifold $M$ embedded in some $\mathbb{R}^n$ and let $f(x) = d(x,M)^2$ be the distance to the manifold squared. Let $V_a$ and $V_b$ be vector fields on $M$, and let $\tilde{V_a}$ and $\tilde{V_b}$ be extensions to vector fields on $\mathbb{R}^n$. We may assume $f$ is smooth on all of $\mathbb{R}^n$ by multiplying by a cutoff function if necessary. 
This book I'm reading claims the following:
"Since the vector fields $\tilde{V}_a$ are tangent to $M$ along the submanifold $M$, a local calculation shows that $\tilde{V}_af$ and $\tilde{V}_a \tilde{V}_b f$ vanish along the rate of $f$. Hence, there is a neighborhood $U$ of $M$ such that $(i) f(x) = 0$ on $U$ if and only if $x \in M$ $(ii)$ for each $R > 0$, there exists $C$ depending on $R$ such that 
$$|\tilde{V}_af(x) | \leq Cf(x) , |\tilde{V}_a \tilde{V}_bf| \leq Cf(x)$$
for all $x \in U \cap B(0,R)$"
Now, I'm convinced about part $(i)$, but I don't see how to do part $(ii)$. Most of the calculations I've done just give me $\tilde{V}_af = 0$. For example, I calculated $f$ in the case that $M = S^2$, $n = 3$. In this case, $f(x) = (1 - (1/||x||))^2$ and any vector field tangent to $M$ would just make $f$ zero since  $f$ only depends on $r$ (If this thought process is wrong, please let me know!). So, can someone help me out please? 
Thanks!
 A: The issue is that either (1) you are calculating the derivative on the manifold $M$ instead of in $U \cap B(0,R)$ which contains points not on $M$ or (2) you are assuming that the extension of the vector fields to $\mathbb{R}^n$ stay "parallel" to $M$ rather than having some radial component (which they may).
I will try to show why the first inequality of (ii) is true.  (Since the proof will be in local coordinates, I apologize for the mess that is about to follow...)
Let $\dim M = m$ and let $k = n-m$.  Fix a point $p$ on $M$ and choose local coordinates $x_1,x_2, \dots, x_m, y_{1}, y_2, \dots,y_k$ based at $p$.  Here the $x_i$ represent the coordinates along the manifold and $y_j$ are the coordinates in the normal direction.  Letting $\partial_{x_i}$ and $\partial_{x_j}$ be the corresponding unit vectors along these coordinates.  Then any vector field $\overline{V}_a$ can be written as
$$\overline{V}_a = \sum_{i=1}^{m} a_i \partial_{x_i} + \sum_{j=1}^{k} b_j \partial_{y_j} = \overline{V}_a^T + \overline{V}_a^N$$
where $\overline{V}_a^T$ is the tangential component and $\overline{V}_a^N$ is the normal component.  Here, $a_i$ and $b_j$ are smooth functions defined on the coordinate neighborhood.  As you noted, $\overline{V}_a^T f = 0$, so we need only deal with the normal component.
In our coordinates, $f = \sum_{j} y_j^2$.  So, $\nabla f = \sum_{j} 2y_j \partial_{y_j}$ and 
\begin{align*} \overline{V}_a^N f &= \nabla f \cdot \overline{V}_a^N \\
& = \sum_{j} 2y_j \partial_{y_j} \cdot \sum_j b_j\partial_{y_j}\\
& = \sum_j 2y_j b_j.
\end{align*}
Applying Cauchy-Schwarz,
\begin{equation}\left| \overline{V}_a^N f\right| \leq C f^{\frac{1}{2}} (\sum_j b_j^2)^{\frac{1}{2}}. \end{equation}
Now, since $\overline{V}_a$ is an extension of $V_a$, we know that $b_j = 0$ for points on the manifold.  In other words, if we expand $b_j$ by a Taylor series centered at $p$.  then, the constant terms are 0, i.e. $$b_j = \sum_k b_{j,k} y_k + \textrm{higher order terms}.$$
That is $|b_j| \leq C \sum_k |y_k|$. So,
$$b_j^2 \leq C \left( \sum_k |y_k|\right)^2 \leq C\sum_k |y_k|^2 = C f.$$
This implies
$$\left| \overline{V}_a^N f\right| \leq C f.$$
