Okay so I think I have an answer for existence of a spatial isomorphism in the case that the representation $\pi_{1},\pi_{2}$ are the identity representaion on $H.$ . Note that the condition $N\vee M'\cong N\overline{\otimes}M'$ spatially is independent of the way $M$ is represented. Indeed, suppose $M$ is represented on $H$ and $U$ is a unitary which conjugates $N\overline{\otimes}M'$ to $N\vee M'.$ If $K$ is another Hilbert spaces, and $M$ is represented as $M\otimes \Bbb C$ on $H\otimes K,$ then $U\otimes 1$ conjugates $N\overline{\otimes}(M'\overline{\otimes}B(K))$ to $N\vee (M'\overline{\otimes} B(K))$ (here $M'$ means the commutant in $H.$) Similarly, suppose we cut the representation by a projection $p$ in $M'.$ Since $U^{*}pU=1\otimes p$ by assumption, the unitary $U(p\otimes p):pH\otimes pH\to pH$ conjugates $N\overline{\otimes}pM'p$ and $N\vee pM'p.$ By the essential uniques of a normal representation of a Von Neumann Algebra, this proves the claim.

Let us first assume $M$ is type $II.$ Since $M$ is type $II,$ I will assume from here on that $M$ is represented on $L^{2}(M,\tau)$ with $\tau$ a (unique up to scalars) semifinite normal trace.

First off, let's note that $M$ cannot be type $II_{1}$ and this isomorphism be spatial, then we have $N'\cap M\cong (N\overline{\otimes} M')'=N'\overline{\otimes} M.$ Assume $M$ is type $II_{1},$ as noted in my last post, $N'\cap M$ cannot be finite dimensional, and being a subfactor of $M$ (being isomorhpic to $N'\overline{\otimes} M$) it must be a $II_{1}$-factor. But the assumption $N'\cap M$ is infinite dimensional implies that the inclusion $N\subseteq M$ is not finite index, i.e. $\dim_{N}(H)=\infty.$ Then, by definition, we know that $N'$ is infinite. (For more details see Section XIX.2 in Takesaki's Theory of Operator Algebras III).

So we may assume $M$ is type $II_{\infty}.$

So if $M$ is a $II_{\infty}$ factor we may focus on the representation of $M$ on $L^{2}(M,\tau)$ with $\tau$ a fixed semifinite normal trace on $M$ (unique up to scalar multiplication). In this case we claim that this isomorphism is spatial if and only if $N'\cap M$ is infinite. Indeed if $N\vee M'\cong N\overline{\otimes}M'$ spatially then taking commutants implies that $N'\cap M\cong N'\overline{\otimes}M$ spatially, in particular since $M$ is infinite so is $N'\cap M.$

Conversely if $N'\cap M=(N\vee M')'$ is infinite, then since $(N\otimes M')'=N'\otimes M$ is $II_{\infty}$ we have two isomorphic Von Neumann algebras with properly infinite commutants. It is known that two such Von Neumann algebras must be spatially isomorphic (see Theorem V.3.1 in Takesaki's Theory of Operator Algebras I).

If $M$ is type $I$ i.e. isomorphic to $B(H)$ we may take the Hilbert space $M$ is represented on to be $H$ with the canonical action of $M.$ In this case since $M'=\Bbb C,$ it is easy to see that $N\cong N\otimes \Bbb C$ spatially.

Of course, one wouldn't really be able to regard this as explicit, because you would have to know how to write one representation of $M$ in terms of another by tensoring with the trivial representation and cutting by projections.