I'll start with the statement of the theorem and its proof, and I'll end by explaining my difficulty understanding the proof. What follows are not Milnor's original words, but rather my best attempt at paraphrasing.
Milnor: Lectures on the h-Cobordism Theorem (free pdf!), Lemma 2.6 (page 9).
Let $(W; V_0, V_1)$ be such that
- $W$ is a smooth, compact $n$-manifold.
- $V_0$ and $V_1$ are smooth submanifolds of $\partial W$.
- $V_0$ and $V_1$ are clopen subsets of $\partial W$.
- $V_0$ and $V_1$ are disjoint, and their union is $\partial W$.
Then there exists a smooth map $f:W \rightarrow [0, 1]$ such that $f^{-1}(0)=V_0, f^{-1}(1)=V_1$, and $df$ is non-zero on an open neighbourhood of $\partial W$.
Proof. Let $U_1, ..., U_k$ be an open cover of $W$ by coordinate charts, each equipped with a diffeomorphic embedding $h_i:U_i \rightarrow \mathbb{R}^n$. We may assume that no $U_i$ meets both $V_0$ and $V_1$, and if $U_i$ meets $\partial W$, then $h_i$ carries $U_i$ diffeomorphically onto the open unit ball intersected with $\mathbb{R}^n_+$. Write $h_i=(x^i_1, ..., x^i_n)$. If $U_i$ meets $V_0$, let $f_i=x^i_n$. If $U_i$ meets $V_1$, let $f_i = 1-x_n^i$. If $U_i$ does not meet $\partial W$, let $f_i = 1/2$ identically on $U_i$. Extend each function $f_i$ to all of $W$ by setting $f_i=0$ outside of $U_i$.
Next, let $\phi_1, ..., \phi_k$ be a partition of unity subordinate to $U_1, ..., U_k$, and let $f=\sum_i \phi_i f_i$. Then $f$ is smooth with $f^{-1}(0)=V_0$ and $f^{-1}(1)=V_1$. It only remains to check that $df\neq 0$ on $\partial W$. Let $q\in V_0$ (respectively, $q\in V_1$). Let $j$ be such that $\phi_j(q) > 0$. Then \begin{align*} \frac{\partial f}{\partial x^j_n} &= \sum_i f_i \frac{\partial \phi_i}{\partial x^j_n} + \phi_i \frac{\partial f_i}{\partial x^j_n}. \tag{1} \end{align*} If $i, l$ are such that $f_i(q)$ and $f_l(q)$ are both non-zero, then $f_i(q)=f_l(q)$ ($=0$ if $q\in V_0$, and $=1$ if $q\in V_1$). Moreover, \begin{align*} \sum_i\frac{\partial \phi_i}{\partial x^j_n} = \frac{\partial}{\partial x^j_n} \sum_i \phi_i = \frac{\partial}{\partial x^j_n} (1)=0. \end{align*} So, \begin{align*} \frac{\partial f}{\partial x^j_n}(q) = \sum_i \phi_i(q) \frac{\partial f_i}{\partial x^j_n}(q). \end{align*} If $q\in V_0$, then $\frac{\partial f_j}{\partial x^j_n}(q)=1$. Furthermore, small variations in $x^j_n(q)$ cannot decrease the value of $f_i(q)$, so $\frac{\partial f_i}{\partial x^j_n}(q) \geq 0$. Hence, \begin{align*} \frac{\partial f}{\partial x^j_n}(q) =\phi_j(q) \frac{\partial f_j}{\partial x^j_n}(q)+ \sum_{i \neq j} \phi_i(q) \frac{\partial f_i}{\partial x^j_n}(q) \geq \phi_j(q)(1) + 0 > 0. \end{align*} So $df \neq 0$ at $q$. A similar argument works if $q\in V_1$.
QED
My confusion: The use of the product rule in equation (1) confuses me. Don't we need $f_i$ to be smooth at $q$ to even discuss $\frac{\partial f_i}{\partial x^j_n}(q)$? Clearly $f_i$ is not necessarily smooth at $q$ if $q\notin U_i$.