Existence of a transversal map prevents density? Let $S$ be a $C^{\infty}$-submanifold of $N$ and suppose that $N-S$ is dense in $N$, where $M,N$ are $m$ and $n$ dimensional $C^{\infty}$-manifolds, respectively.
In this post the answering poster makes the following arguments:
If $S$ is of codimension $\leq m$ in $N$, then exists $C^{\infty}$ smooth map $f:M \to N$ which intersects $S$ transversely in at least one point.  So $f$ cannot be in the $C^0$-closure of $C(M,N')$, where $N'=N-S$.
Why is this true?  I've never had to use any intersection theory before so the argument has left me at a loss of direction.
I've been reading and this may have some link with this result?  But its still not clear to me.
 A: The terminology is a bit odd: One says that two submanifolds intersect transversally or that a map is transversal to a submanifold. The terminology that a map intersects a submanifold (transversally) is nonstandard. 
Option 1: What's meant is that there exists $p\in M$ such that $f(p)\in S$ and $f$ is transversal to $S$ at $p$. 
Option 2: What's meant is  that there exists a point $q\in S$ such that $q\in f(M)$ and $f$ is transversal to $S$ at every $p\in f^{-1}(S)$. 
The existence of $f$ holds regardless, just one needs to do more work in the case of Option 2. 
The first step is simple linear algebra. Suppose that $W$ is a linear subspace of ${\mathbb R}^n$ whose codimension is $\le m$. Then there exists an $m$-dimensional linear subspace $V\subset {\mathbb R}^n$ transversal to  $W$. (Their intersection contains the origin, of course.)  
Now, since $S$ is a submanifold in $N$, we can take a point $q\in S$ and local chart
$$
\psi: U\subset N\to {\mathbb R}^n
$$
such that $\psi(S\cap U)$ is an open neighborhood of $0$ in a linear subspace $W\subset {\mathbb R}^n$ and $\psi(q)=0$. 
Then $W$ has codimension at most $m$. 
Take $V\subset {\mathbb R}^n$ as above. Since $M$ is $m$-dimensional, there exists a local chart $\phi: M\to V$ sending the given $p\in M$ to $0$. Extend $\phi$ to a smooth map $\eta: M\to V\cap \phi(U)$ (after shrinking the open neighborhood of $p\in M$ appropriately). Now, compose $\eta$ with $\phi^{-1}$ and obtain a smooth map $f: M\to U$ 
which is transversal to $S$ at $p$. This takes care of Option 1. 
To deal with Option 2, one first the following lemma (I leave you a proof as an exercise): 
Lemma. Let $M$ be an $m$-dimensional manifold. Then there exists a smooth map $\theta: M\to D^m$ to the open disk $D^m\subset V={\mathbb R}^m$, such that $0\in \theta(M)$ and $0$ is a regular value of $\theta$. 
Use this map $\theta$ as $\eta$ in the above proof. 
Edit.
Proposition.  Suppose that $f: M\to N$ satisfies Option 1. Then $f$ does not belong to the closure of $C(M,N')$ equipped with topology of uniform convergence on compacts, where $N'=N-S$.
Proof. By density of smooth maps, it suffices to show that $f$ cannot appear as a limit (uniform on compacts) of a sequence of smooth maps $f_i: M\to N'$.  (Here I equip $N$ with some auxiliary Riemannian metric.) 
The proof is a variation on the standard transversality arguments which you can find, for instance, in "Differential Topology" by Guillemin and Pollack. 
Pick a submanifold $M'\subset M$ of dimension equal to codimension of $S$ in $N$, containing $p$ and such that $f: M'\to N$ is transversal to $S$ at $p$. Restricting $f$ (and a sequence $(f_i)$) to $M'$, the problem is reduced to the case when $m=dim(M)=codim_N(S)$. 
Let $U$ as above be a coordinate neighborhood of $q=f(p)$ in $N$; we can assume that the closure of $U$ is diffeomorphic to a closed $n$-ball in ${\mathbb R}^n$.  Then (by taking $U$ sufficiently small) we can also assume that there exists a compact codimension 0 submanifold with smooth boundary $\Omega\subset M$ such that $f(\partial \Omega)\subset \partial D$, 
$f(\Omega)\subset D$ and 
$$
\{p\}= f^{-1}(S)\cap \Omega. 
$$
For all sufficiently large $i$, each map $f_i|\Omega$ are homotopic to $f|\Omega$ by the "straight-line homotopy" $H_i: \Omega\times [0,1]\to N$ with $H_i(x,0)=f(x), H_i(x,1)=f_i, x\in \Omega$. 
For the sake of a contradiction, I will assume that $f_i(\Omega)\cap S=\emptyset$.  Then, WLOG (by replacing the straight-line homotopy by its small smooth perturbation), we can assume that each $H_i$ is transversal to $S$. In particular, each 
$$
H_i^{-1}(S)
$$
is a compact 1-dimensional submanifold $s_i$ in $\Omega\times [0,1]$. 
When $i$'s are sufficiently large, $H_i(\partial \Omega \times[0,1])\cap S=\emptyset$. Therefore, the boundary of $s_i$ is contained in $int(\Omega)\times \{0\}$. This boundary is nonempty since $p\in \partial s_i$. Hence, there exists  $p'\in \Omega -\{p\}$ such that $f(p')\in S$. A contradiction. qed    
