The answer is 'yes, there always is such a vector field $X$' and, in particular, the answer to your 'new question' is also 'yes'. (In fact, the first 'yes' implies the second 'yes', but the second 'yes' is used in the proof of the first 'yes'.)

Here is an outline of the proof (and it works also in the codimension $1$ case; I'm not sure why you want to exclude that):

First, since things are local around $N$, you can reduce to the case that $M$ is a vector bundle and $N$ is the zero section of that vector bundle (in which case, we can obviously identify $N$ with the base of the vector bundle). This is a standard result in differential topology, and one can find proofs in many places. Basically, you fix a Riemannian metric on $M$ and use the normal exponential map to show that a neighborhood of $N$ in $M$ is diffeomorphic to a neighborhood of the zero section of the normal bundle to $N$.

So suppose that $M^{n+k}$ is a vector bundle over $N^n$ and that $\mu$ is a volume form on $M$. Let $R$ be the vector field on $M$ whose time $t$ flow is scalar multiplication by $e^t$ in the fibers of $M\to N$ (which are vector spaces, so scalar multiplication is well-defined). Note that $R$ vanishes along $N$.

Then I claim that, for $k\ge 1$, there is a unique function $f$ on $M^n$ such that the vector field $X = fR$ has divergence $1$ with respect to $\mu$, i.e., that $\mathcal{L}_X\mu = \mathrm{d}\bigl(f\,\iota_{R}\mu\bigr) = \mu$, where $\iota_R\mu$ is the $(n{+}k{-}1)$-form that is the 'interior product' (aka 'left hook') of $R$ with $\mu$.

This is a claim about the unique solvability of a singular, linear first-order PDE: Let $\mathrm{d}\bigl(\iota_{R}\mu\bigr) = \kappa\,\mu$. Then $\kappa$ is a smooth function on $M$ that satisfies $\kappa(p) = k>0$ for all $p\in N$. I am claiming that there exists a unique smooth (and positive) function $f$ on $M$ that satisfies the linear inhomogeneous equation
$$
\mathrm{d}f(R) + \kappa\,f = 1.
$$
Since $R$ vanishes along $N$, this implies $f(p) = 1/k$ for all $p\in N$.

You can see what makes this a little delicate by examining what happens along a single line in a single fiber: If $v\in M_p$ is a nonzero element in the fiber over $p$, we can parametrize the line $\mathbb{R}v\subset M_p$ by $t\mapsto t{\cdot}v$ for $t\in \mathbb{R}$. In this case, the PDE to be solved becomes an ODE
$$
tf'(t) + h(t)\,f(t) = 1
$$
where $h$ is a smooth function on $\mathbb{R}$ that satisfies $h(0) = k>0$. This is a regular, singular ODE, and while its (unique) smooth solvability is a classical fact, this seems not to be that well-known these days. The uniqueness on each line tells you that there is at most one smooth solution $f$ to the full equation on $M$, and it's not hard to use the uniqueness to show that, in fact, you get global smoothness for $f$ as well.

If you are doubtful, I can supply the proof. While this is a classical ODE/PDE fact, I don't, off-hand, remember a good source, just a proof (which must be the standard one).

**Added remark about ODE/PDE fact:** First, consider the ODE on the real line $\mathbb{R}$
$$
t\,f'(t) + h(t) f(t) = g(t)
$$
where $h$ and $g$ are given smooth functions on $\mathbb{R}$ and $h(0) = k>0$. We want to show that there is a smooth solution $f$ and that it is unique. Write $h(t) = k - t m'(t)$ for some smooth function $m$ on the real line. When $g = 0$, the only solutions are $f(t) = c\,t^{-k}\mathrm{e}^{m(t)}$, where $c$ is a constant, and hence only when $c=0$ is the solution smooth. For general $g$, we use variation of parameters and look for a solution of the form $f(t) = c(t) t^{-k} \mathrm{e}^{m(t)}$
for some function $c(t)$ that vanishes to order at least $k$ at $t=0$. Substituting this into the above equation, we find that $c'(t) = t^{k-1}g(t)\mathrm{e}^{-m(t)}$, so, since we want $c(0)=0$, we have
$$
c(t) = \int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau.
$$
Since $k>0$, this integral vanishes to order $k$ at $t=0$. Thus,
$$
f(t) = \mathrm{e}^{m(t)}t^{-k}\int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau
$$
is a smooth solution. It is unique because we have already shown uniqueness when $g=0$. For use below, note that, if $g$ vanishes to order $l>0$ at $t=0$, then so does $f$, and we always have $f(0) = g(0)/k$.

Now, returning to the general case of $\mathrm{d}f(R) + \kappa f = 1$,
we see, by applying the above argument to each line in $M_p\subset M$
for $p\in N$, that there is a unique function $f$ on $M$ that satisfies this equation and that $f$ is smooth on $M$ except possibly along the zero section $N\subset M$ itself. It is clearly smooth on every line through $0_p\subset M_p$, but one might worry that it is not smooth on a neighborhood of $N$ in $M$.

However, the following argument shows that this is not the case: Let $F_k(M,N)$ denote the space of smooth functions on $M$ that vanish to order at least $k$ along $N$. This is a decreasing filtration of $C^\infty(M) = F_0(M,N)\supset F_1(M,N)\supset\cdots$. One easily sees that the linear operator $D(f) = \mathrm{d}f(R) + \kappa\,f$ maps $F_i(M,N)$ into $F_i(M,N)$ and hence induces a linear operator $D_i:G_i(M,N)\to G_i(M,N)$ on the associated graded
$$
G_i(M,N) = F_i(M,N)/F_{i+1}(M,N)\simeq S^i(M^*)
$$
(bearing in mind that $M$ is a vector bundle over $N$). Since $R$ is the Euler (radial) vector field on each fiber $M_p$ and since $\kappa(p)=k$ for all $p\in N$, it follows that $D_i$ is simply multiplication by $i{+}k$ for $i\ge0$ and hence is an isomorphism of $G_i(M,N)$ with itself for $i\ge0$.

Consequently, for every $i\ge0$, there exists a smooth $f_i$ on $M$
such that $\mathrm{d}f_i(R) + \kappa\,f_i = 1 - h_i$ where $h_i\in F_{i+1}(M,N)$. Using the above 'integration construction on each line', we can find a function $u_i$ on $M$ that is smooth away from $N$, vanishes to order $i{+}1$ along $N$ and satisfies $\mathrm{d}u_i(R) + \kappa\,u_i = h_i$. It follows that $f = f_i + u_i$. Thus, for every $i\ge0$, $f$ can be written as the sum of a function $f_i$ that is smooth on $M$ and a function $u_i$ that vanishes to order $i{+}1$ along $N$ and is smooth away from $N$. Consequently, $f$ is differentiable to all orders along $N$ and hence is smooth on all of $M$ as desired.

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