Notion of linking between two general $p$ and $q$ manifolds embedded in a higher dimensional manifold

Sure, there are ways to make sense of linking beyond the constraints you mention.

I suppose the most basic would be: two disjoint submanifolds $A$ and $B$ of a manifold $M$ are unlinked if you can find disjoint embedded $m$-dimensional discs $D_1, D_2 \subset M$ such that $A \subset D_1$ and $B \subset D_2$. Here $m=dim(M)$.

The manifolds would be linked if they are not unlinked.

A nice feature of this definition is you can talk about linking of three or more objects, so you can similarly talk about Brunnian properties, i.e. $k$ objects can be linked but any collection of $k-1$ sub-objects are unlinked.

For example, there is a classical theorem of Schubert that says if you have two (or more) disjoint non-trivial knot exteriors in $S^3$, then they are unlinked. By knot exterior I mean a compact 3-manifold diffeomorphic to the complement of an open tubular neighbourhood of a knot in $S^3$. As you likely know, if the knot exterior is trivial it would be diffeomorphic to $S^1 \times D^2$, and such objects can be linked in $S^3$.

I believe this is a commonly-used notion of linking that generalizes the definition you mention. Using this definition, linking of surfaces in $\mathbb R^3$ or $S^3$ is a fairly easy definition to work with, once you are comfortable with the tools of 3-manifold theory.

In your question you refer to "the linking between $M$ and $N$" but really you are talking about linking numbers. Interesting numerical linking invariants can take a fair bit of work to define, and John Klein's answer is directed towards that kind of question. But if all you care about is what the term "linking" means, that is much more elementary, and what my answer is directed towards.

edit: here are more details to support my comment below. This argument is typical of Hassler Whitney's work. I'm not certain if he ever made this specific argument, but the general mode of argument very much is due to him.

For convenience, let's say $M$ and $N$ are compact submanifolds of $\mathbb R^k$. We can get by without this assumption, but it makes the arguments less technical.

By compactness, there is some vector $v \in \mathbb R^k$ so that $M+v$ is disjoint from $N$. Consider the function

$$F: [0,1] \times M \to [0,1]\times \mathbb R^k$$

given by $F(t,p) = (t, tv+p)$.

Thus $F(0,M) = \{0\} \times M$ and $F(1,M) = \{1\} \times (v+M)$.

So $F([0,1] \times M)$ is disjoint from $[0,1] \times N$ at the start and the end of the interval, but maybe not in the interior.

We can perturb $F$ a little bit (not affecting its values at the endpoints) to make it transverse to $[0,1]\times N$. Provided $k+1 > (m+1)+(n+1)$ the transverse map would be disjoint from $[0,1]\times N$, i.e. $k \geq m+n+2$.

Since the transversal perturbation can be made arbitrarily small in the $C^\infty$-topology, we can assume it represents an isotopy of $M$. Thus $M$ can be isotoped to be separable from $N$ by disjoint embedded discs. So by isotopy-extension, $M$ and $N$ can be separated by disjoint embedded discs.

There are various approaches to this. One approach, developed by Bruce Williams and me, uses homotopy theory. See:

Homotopical intersection theory, I. Geometry & Topology 11, (2007) 939–977 arXiv: math/0512479

For a multirelative generalization, see:

Homotopical Intersection Theory, III: multi-relative intersection problems. Algebraic & Geometric Topology 19-3 (2019), 1079--1134. arXiv:1212.4420

The basic idea is is not hard to explain. Suppose $P, Q$ and $N$ are manifolds. Then there is a commutative diagram $\require{AMScd}$ $$ \begin{CD} \text{map}(P,N) \times \text{map}(Q,N) \setminus \cal D @>>> \text{map}(P\times Q,N\times N \setminus \Delta)\\ @VVV @VVV \\ \text{map}(P,N) \times \text{map}(Q,N) @>>> \text{map}(P\times Q,N\times N) \end{CD} $$ where: the vertical maps are inclusions of function spaces, $\cal D$, the discriminant, is the subset of pairs $f: P\to N$, $g: Q\to N$ such that $f(P) \cap g(Q)$ is nonempty, and the horizontal maps take a pair $(f,g)$ to the product map $f\times g$. Here, $\Delta$ is the diagonal subspace of $N\times N$.

Given a basepoint $(f,g) \in \text{map}(P,N) \times \text{map}(Q,N)$, let $$ F_{f,g} $$ denote the homotopy fiber of the left vertical map. Similarly, let $L_{f,g}$ denote the homotopy fiber of the right vertical map taken at $f\times g$. Then the induced map $$ F_{f,g} \to L_{f,g} $$ is is a kind of generalized linking invariant.

In the first cited paper, Bruce and I identify the homotopy type of $L_{f,g}$ in a range. The answer is that there is a map $$ L_{f,g} \to \Omega^{\infty+1} E(f,g)^{\xi} $$ where $E(f,g)$ is the homotopy pullback of the maps $f$ and $g$ and $\xi$ is a certain virtual vector bundle with which we form the Thom spectrum $E(f,g)^{\xi}$. The displayed map was shown to be $(2n-p-q-3)$-connected (where the lower case letters correspond to the dimensions of the manifolds in question).

The "generalized linking number" is then the induced homomorphism on path components $$ \pi_0(F_{f,g}) \to \Omega_{p+q-n+1}(E(f,g);\xi) $$ where the target denotes the bordism group of $E(f,g)$ twisted by $\xi$ in dimension $p+q-n+1$.

Perhaps to most interesting case occurs when $N$ is euclidean space $\Bbb R^n$, since $F_{f,g} $ has the homotopy type of the space of link maps $(f,g): P\amalg Q \to \Bbb R^n$ in that case. If in addition $P$ and $Q$ are stably parallelizible, then the homomorphism reduces to the form $$ \pi_0(F_{f,g}) \to \pi^{\text{st}}_{p+q-n+1}(P\times Q)\, , $$ where the target is the stable homotopy group of $P\times Q$ in the displayed dimension. Different forms of this invariant, in the case of spheres, were investigated my several authors (Koschorke comes to mind).

For another, more geometric approach, I recommend the papers of Brian Munson (at least one of which is joint with Tom Goodwillie).