Module isomorphism and length of tensor product.

In the paper your image refers to, $\Psi_A(M)$ is defined as $\hat{\Psi}_A(M):=\frac{M}{M[I_A]+I_AM}$.
The first part of your question is why do we have an isomorphism $\hat{\Psi}_A(M) \cong \Psi_A \otimes_{\mathcal{O}}\frac{M}{M[I_A]}$. To see this, first observe that we have $\frac{A}{I_A} \otimes_A \mathcal{O} \cong \mathcal{O}/\lambda(I_A)=\Psi_A$. Indeed, this is just a special case of the basic isomorphism $ R/J \otimes_R S \cong S/JS$ valid for any commutative ring $R$, ideal $J \subseteq R$ and $R$-algebra $S$). Therefore
$$\Psi_A \otimes_{\mathcal{O}}\frac{M}{M[I_A]}= \frac{A}{I_A} \otimes_A \mathcal{O} \otimes_{\mathcal{O}}\frac{M}{M[I_A]}=\frac{A}{I_A} \otimes_A \frac{M}{M[I_A]}$$
The latter is isomorphic to $\frac{\frac{M}{M[I_A]}}{I_A \cdot (\frac{M}{M[I_A]})}$ again by a basic isomorphism for tensor products analogous to the one I mentioned above. Using basic isomorphisms regarding quotients, this last object is identified with $\frac{M}{M[I_A]+I_AM}$. Putting all this together, we have justified the isomorphism $\Psi_A \otimes_{\mathcal{O}}\frac{M}{M[I_A]} \cong \hat{\Psi}_A(M)$.

Now, to answer your second part of the question regarding the identity of lengths, we use the shorthand notation $l$ to mean length $\mathcal{O}$. Then we have to show that $l(\hat{\Psi}_A(M))=l(\Psi_A) \cdot d$ where $d$ is the rank of $\frac{M}{M[I_A]}$ which is indeed a free module of finite rank by Lemma 3.2 in the cited paper, i.e $\frac{M}{M[I_A]} \cong \mathcal{O}^d$. Thus
$$\Psi_A \otimes_{\mathcal{O}}\frac{M}{M[I_A]} \cong \Psi_A \otimes_{\mathcal{O}} \mathcal{O}^d \cong \Psi_A^d$$
where the last isomorphism holds by standard properties of the tensor product. Finally, the length of $\Psi_A^d$ is $d$ times the length of $\Psi_A$, by standard properties of the length (it is additive in direct sums of modules). Thus we are done.