I think the following works, but it is ad hoc. I would have suspected Hirsch had something more general in mind.

Because there is a tubular neighborhood of the embedding $f$ in $N$, we can work in an open subset of the zero section of the normal bundle $N_f(M)$. Because of this, I'll identify the points $f(x)$ with $x$ in the normal bundle. The open sets $A_i$ and $B_i$ can be taken to be of the form $\pi^{-1}(A'_i)$ and $\pi^{-1}(B'_i)$ where $A'_i$ and $B'_i$ are open subsets in $M$, with $M\setminus U_i \subseteq B'_i$ and $K_i \subseteq A'_i$.

We want to find $\epsilon$'s about each $x \in X$ so that all points in $B(x; \epsilon)$ in the normal bundle stay in $A_i$, whenever $x \in K_i$, and stay in $B_j$ whenever $x \in M \setminus B'_j$.

First step: If $x \in K_i$, there are only finitely many $K_j$ that intersect $K_i$ (if there were infinitely many, we can take a sequence of elements of $K_i$ that lie in some other $K_j$, this has a convergent subsequence with limit in $K_i$, but then this limit point lies in infinitely many other $K_j$, which is prohibited by local finiteness of the cover). So we can now measure the distance to any $x \in K_i$ to $N \setminus A_i$ to get $\epsilon_i$. We require that $x$ can't move more than $\epsilon_i$ then, but we must impose this condition for every $j$ that $x \in K_j$. We just showed there are only finitely many $j$ though, so there is a smallest positive of $\epsilon$ for each $x$.

Second step: Suppose now that $ x \in K_j$ and $x \in X \setminus U_i$. We can then measure the distance from $x$ to $N\setminus B_i$, call it $\epsilon_{i,j}(x)$ which is positive. We require that for a fixed $x$, for every $j$ and $i$ for which $x \in (X \setminus U_i) \cap K_j$ that we do not move from $x$ more than $\epsilon_{i,j}(x)$. The worry is that there is no positive $\epsilon(x)$ that is smaller than all of the $\epsilon_{i,j}(x)$. Let us check if this can happen.

We know that each $x$ lies in only finitely many $K_j$, but $x$ can lie in infinitely many of the sets $X \setminus U_i$. So for each $j$, consider if it is possible that there is a sequence of $i_k$ so that $x \in \setminus U_{i_k}$ with the distance $\epsilon_{j,i_k} \to 0$ as $k \to \infty$. If this were the case, then there would be a sequence of points $z_{i_k}\in N \setminus B_{i_k}$ in the normal bundle, lying over points $y_{i_k} \in M \setminus B'_{i_k}$ so that the distance $d(x, y_{i_k}) \to 0$ as $k \to \infty$. But then, as $x \in K_j \subseteq \mathrm{Int}(U_j)$, there are infinitely many values of $k$ for which $y_{i_k} \in U_j$. That is, local finiteness at $x \in K_j$ is contradicted. We therefore conclude that each $x \in X$ admits an $\epsilon(x)$ for which all points $z \in B(x; \epsilon(x))$ lie in all possible $N \setminus B_i$.

To finish up the argument, we just need to pick the smallest value of the $\epsilon$'s determined above on each compact set $K_j$. This then gives an open set $\mathcal{N}_1$ in the strong Whitney topology $C^0_S(M, N)$ for which all $g$ in this set satisfy $g(K_j) \subseteq A_j$ and $g(M \setminus U_j) \subseteq B_j$ for all $j$.