If your interpretation of "smooth submanifold" is smooth embedded submanifold (meaning that it has the subspace topology), then these two definitions are equivalent. The proof of the existence of a holomorphic slice works just like the smooth case, but using the holomorphic version of the inverse function theorem instead of the smooth one. Here's a sketch of the proof.
Suppose $N\subseteq M$ is a smooth embedded submanifold with a complex structure, and let $\iota\colon N\to M$ denote the inclusion map. Given $p\in N$, we can choose holomorphic coordinates on a neighborhood $V$ of $p$ in $N$, and holomorphic coordinates on a neighborhood $W$ of $\iota(p)$ in $M$. Since $N$ has the subspace topology, after shrinking $V$ and $W$ if necessary we can assume that $V=W\cap \iota(N)$. Since the question is local, we might as well replace $M$ and $N$ by $V$ and $W$, identified with open subsets of $\mathbb C^r$ and $\mathbb C^n$, respectively. Then $d\iota_p$ is an injective complex-linear map from $\mathbb C^r$ to $\mathbb C^n$. After a complex-linear change of coordinates, we may assume that the image of $d\iota_p$ is the span of the first $r$ standard coordinate vectors. Let $Y$ be the span of the complementary $n-r$ coordinate vectors, and define $\Phi\colon \mathbb C^n = \mathbb C^r\times Y \to \mathbb C^n$ by $\Phi(x,y) =\iota(x)+y$. The holomorphic inverse function theorem shows that $\Phi$ is a biholomorphism onto some neighborhood $U$ of $(p,0)$, and $\Phi^{-1}$ is the required local holomorphic chart.
Assume $M$ is a compact $n$-manifold and that $\xi$ is a $n$-plane bundle on $M$. By transversality, there exists a section $s$ of $\xi$ that has finitely many isolated zeros. We shall "push" each zero onto the special point $x$.
For each zero of the section $s$, there is an embedded arc connecting it to $x$ since $M$ is connected. Take a small open neighborhood $U$ of this arc such that $\bar{U}$ does not contain any other zeros of $s$. If $U$ is chosen small enough, it will be contractible and $\xi|_U$ is trivial; we use this to identify all the fibers of $\xi|_U$. We claim that there is a smooth modification $\tilde{s}$ of $s$ such that $\tilde{s} \equiv s$ outside of $U$, and the only zero of $\tilde{s}$ inside $U$ is at $x$.
Without loss of generality, by applying appropriate diffeomorphisms, we may assume that $U$ is the unit ball $B^n$ in $\mathbb{R}^n$ and that $x = 0 \in B^n$. We define $\tilde{s}$ to be identical to $s$ outside of $U$, and inside of $U \cong B^n$, $$\tilde{s}(r, \theta) = \rho(r) \cdot s|_{\partial B^n}(\theta),$$ where $\rho: [0,1] \to [0,1]$ is a smooth function such that $\rho \equiv 1$ in a neighborhood of $1$, $\rho(r) = e^{-1/r}$ in a neighborhood of $0$, and $\rho(r) = 0$ iff $r = 0$. Then the only zero of $\tilde{s}$ inside $U$ is at $x$, as desired.
Repeating these modifications for each zero of $s$, we obtain a section whose only possible zero is located at $x$.
It turns out that this procedure of pushing (and possibly merging) zeros of sections is very useful.
For example, suppose now that $M$ has nonempty boundary $\partial M$. Attach a collar to the boundary (which does not change the topology of $M$), construct a section with isolated zeros, and push all of them onto the collar as above. Then simply detach the collar to get a section with no zeros.
If $M$ is noncompact, take a compact exhaustion $\emptyset = K_0 \subset K_1 \subset K_2 \subset \cdots \subseteq M = \bigcup_i K_i$. Given zeros in $K_i \setminus K_{i-1}$, we push them all to $K_{i+1} \setminus K_i$. Observe that this process leaves the section defined on $K_{i-1}$ unchanged, so as we keep pushing all the zeros of a section off to infinity, we obtain a well-defined nonvanishing section of $\xi$.
For full disclosure, I'm not sure whether the given construction of $\tilde{s}$ is the most efficient, and checking that it is indeed smooth at the center is a bit of a hassle.
Also, there is a simple argument to show that nonvanishing sections exist on noncompact manifolds if you know a bit about Euler classes. The only obstruction to a nonvanishing section of a rank $n$ bundle on a $n$-manifold $M$ is the Euler class, which is necessarily zero since $H^n(M) = 0$.
Best Answer
Right, so I figured it out thanks to Philip Andreae's comments. For each $p\in S$, pick a chart $(U_p,\phi_p)$ in $M$ such that $U_p\cap S\subset U_p$ is a $k$-slice. Let $\sigma:S\to E|_S$ be the section. Then $\sigma$ restricts to a section on $U_p\cap S$. If $W_p$ is the associated smooth local trivialization on $M$ (i.e. $\Phi:\pi^{-1}(W_p)\to W_p\times\mathbb{R}^n$) around $p$, after replacing $U_p$ with $U_p\cap W_p$, we can assume that the smooth local trivialization exists around $U_p$. Then there is a local frame $\tau_1,\dots,\tau_n$ associated with $U_p$. Thus $\sigma=(\sigma^1\tau_1,\dots,\sigma^n\tau_n)$ for $\sigma^i:U_p\cap S\to \mathbb{R}$ smooth. Since $U_p\cap S\subset U_p$ is closed and $U_p\subset M$ is a submanifold, we can extend $\sigma^i$ to $\tilde{\sigma}^i:U_p\to \mathbb{R}$ in the canonical way. Thus we have extended $\sigma$ on $U_p\cap S$ to $\sigma_p:U_p\to E|_{U_p}$ expanded in local coordinates. Now we do the usual business using partitions of unity to create some $\tilde{\sigma}:\bigcup_{p\in S} U_p=U\to E|_U$, which, by a standard calculation, extends $\sigma$ and is a smooth section.
If $S\subset M$ is properly embedded, then we can expand the partition of unity argument to have $\tilde{\sigma}:M\to E$ a global smooth section.