Do Betti numbers beyond the first have a "number of cuts" interpretation? I have heard stated the following
Theorem. If $\Sigma$ is a (orientable) surface, then $\mathrm b_1(\Sigma)$ counts the maximum number of "circular cuts" (embedded circles $C_1,\ldots,C_m$) that you can make on $\Sigma$ without disconnecting it (i.e. with $\Sigma\smallsetminus (C_1\cup\ldots\cup C_m)$ still being connected).

Is there a similar interpretation also for higher Betti numbers $\mathrm b_k(M)$ of manifolds in general?

(I've been asked this by a non-mathematician, but I think it fits MO. If the topologists here think it's too trivial or well known, please move it to MSE).
Edit1: It wouldn't be bad to know if the above "theorem" is actually a theorem, and a reference to a rigorous proof. 
Edit2: In the above, by a "generalization" I mean a statement of the form: $b=b_k(M)$ counts the maximum number of embedded orientable submanifolds $N_1,\ldots,N_b$ such that $M\setminus (N_1\cup\ldots\cup N_b)$ has clear topological property $\boldsymbol{\mathrm{P}}$. Maybe property $\boldsymbol{\mathrm{P}}$ could be about some $(k-1)$-connectedness condition? For the $b_1$ case, this would involve $0$-connectedness, i.e. just connectedness, and this is the case for $\dim M=2$ if the theorem I quoted above is true.
 A:  Edit.  I misunderstood the question, so what follows may not be relevant, but it may be useful anyway.
The cut number $c(M)$ of a manifold $M$ is usually defined as the maximum number of disjoint two-sided codimension-1 submanifolds whose complement is connected. By collapsing such a configuration to a graph, one sees that $\pi_1(M)$ surjects onto the free group with $c(M)$ generators. This implies that $H_1(M)$ surjects onto the abelian group of rank $c(M)$, and therefore
$$b_1(M) \geq c(M).$$
However, it happens that these two numbers are often distinct. See for instance this paper of Harvey.
(Collapsing to a graph means that you pick disjoint tubular neighbourhoods $S\times [-1,1]$ of the connected two-sided submanifold $S\subset M$, then you collapse the complement of all these tubular neighbourhoods to a point $p$, and also each factor $S\times [-1,1]$ to an arc $[-1,1]$ with endpoints in $p$.)
A: Here is one way that your observation about the first Betti number generalizes to higher-dimensional manifolds.
Suppose that $M$ a compact, orientable smooth manifold and $W$ is a regularly embedded, closed, orientable submanifold of codimension 1. Then if the number of path components of $W$ is greater than the first Betti number $b_1(M)$, then $M \setminus W$ has more path components than $M$.

Here's the proof. Consider the long exact sequence
$$
\dots \to H_1(M \setminus W) \to H_1(M) \to H_1(M, M \setminus W) \to H_0(M \setminus W) \to H_0(M) \to H_0(M, M \setminus W) \to 0.
$$
Since $W$ is regularly embedded, it has a neighborhood diffeomorphic to the (1-dimensional) normal bundle $\nu$ of the embedding. By the excision axiom, there is an isomorphism
$$
H_k(M, M \setminus W) \cong H_k(\nu, \nu \setminus W).
$$
(Here we view $W$ as the same as the zero-section of $\nu$.)
Since $W$ and $M$ are both orientable, the normal bundle is orientable. Since the normal bundle is orientable of dimension 1, there is a Thom isomorphism
$$
H_k(\nu, \nu \setminus W) \cong H_{k-1}(W).
$$
Therefore, part of our exact sequence becomes
$$
\dots \to H_1(M) \to H_0(W) \to H_0(M \setminus W) \to H_0(M) \to 0.
$$
The rank of $H_0(W)$ is the number of path components of $W$, so if $W$ has more path components than $b_1(M)$ then the map $H_1(M) \to H_0(W)$ cannot possibly be surjective. If it is not surjective, then the image of $H_0(W)$ in $H_0(M \setminus W)$ is nontrivial. Since this image is the kernel of the next map, the map $H_0(M \setminus W) \to H_0(M)$ is not injective. But this map can only fail to be injective if there are more path components in $M \setminus W$ than there are in $M$.

Here are some observations about this proof.


*

*The difference between the number of path components of $W$ and $b_1(M)$ actually gives us a lower bound on the number of new path components in $M \setminus W$, and if $b_1(M) = 0$ we get an exact count.

*If we toss out the word "orientable" everywhere but replace homology with mod-2 homology everywhere, we get a similar bound using the mod-2 Betti number. This sees that we can remove the nonorientable submanifold $\Bbb{RP}^2 \subset \Bbb{RP}^3$ and still have a connected space, even though this can't happen for orientable submanifolds of $\Bbb{RP}^3$.

*As given, the argument is only in one direction. It gives no guarantees that we can find an oriented submanifold $W$ with $b_1(M)$ connected components so that $M \setminus W$ is still connected.

*If we replace codimension 1 in this argument with codimension $r$, we instead get an exact sequence
$$
\dots \to H_r(M) \to H_0(W) \to H_{r-1}(M \setminus W) \to H_{r-1}(M) \to 0.
$$
By the same argument, this allows us to say: if $b_r(M)$ is less than the number of connected components of $W$, the map $H_{r-1}(M \setminus W) \to H_{r-1}(M)$ cannot be injective. This is not quite as punchy as the statement for path components, but the de Rham theorem tells us that it does still indicate something about the existence of new differential forms on $M \setminus W$.

*I don't think, in the original problem, that you wanted the submanifolds to be disjoint. If instead we have connected, closed, regularly embedded, orientable submanifolds $W_1,\dots,W_d$ of codimension $k$ which intersect transversely, one can show by a more complicated inductive argument with the (relative) Mayer-Vietoris sequence that $H_k(M, M \setminus \cup W_i)$ has rank at least $d$, and find that the map $H_{d-1}(M \setminus \cup W_i) \to H_{d-1}(M)$ has kernel of rank at least $d - b_k(M)$.
