Let $\mathcal{C}$ be some small category and consider the representable functor
$$\text{Hom}(,N):\mathcal{C}\rightarrow \text{Ab}.$$
The functor
$$-\otimes_{\mathbb{Z}} M:\text{Ab}\rightarrow \text{Ab}$$
is representable if and only if $N$ is a finitely generated projective $\mathbb{Z}$-module (see When is $M \otimes_A -$ representable?). The composition of these functors
$$\mathcal{C}\rightarrow \text{Ab}, \quad X\mapsto \text{Hom}(X,N)\otimes_{\mathbb{Z}}M$$
is thus representable if $M$ is a finitely generated projective $\mathbb{Z}$-module, as the composition of representable functors is representable. Hence $M$ being a finitely generated projective $\mathbb{Z}$-module is a sufficient condition for $\text{Hom}(-,N)\otimes_{\mathbb{Z}}M$ to representable. Is it also necessary?
Composition of representable functors
category-theoryfunctorsrepresentable-functor
Best Answer
It's not actually true that (co)representables are always closed under composition. Fix a Bénabou cosmos $\mathcal V$ (e.g., $\mathbf{Set}$ or $\mathbf{Ab}$) and let $\cal C$ be a $\cal V$-category, $\mathcal C(a,-):\mathcal C\to\mathcal V$ and $\mathcal V(v,-):\mathcal V\to\mathcal V$ be (co)representable functors. If the composite $\mathcal V(v,\mathcal C(a,-))\cong\mathcal C(v\odot a,-)$ is also (co)representable, then the (co)representing object $v\odot a$ is precisely the copower of $a$ by $v$. Therefore, composites of (co)representables are only (co)representable when $\mathcal C$ is copowered over $\mathcal V$.
Therefore, given your (necessarily $\mathbf{Ab}$-enriched) category $\mathcal C$, we have to assume it is copowered over $\mathbf{Ab}$ or else your composite will not be representable even if $M$ is finitely generated and projective. Moreover, you should note that your functor $\operatorname{Hom}(-,N)$ is contravariant in $\mathcal C$. To remedy this, I will replace $\mathcal C$ by $\mathcal C^{\mathrm{op}}$ and work instead with trying to determine the (co)representability of $$ X\mapsto \mathcal C(N,X)\otimes_{\Bbb Z}M $$ As you mentioned, if $M$ is finitely generated and projective, then $(-)\otimes_{\Bbb Z}M$ will be corepresented by $M^* := \operatorname{Hom}_{\mathbf{Ab}}(M,\Bbb Z)$, so if $\mathcal C$ is copowered over $\mathbf{Ab}$, then the above functor will be corepresented by $M^*\odot N$.
In the converse direction, there are instances where $M$ will necessarily be finitely generated and projective. For instance, take $\mathcal C := \mathbf{Ab}$ and $N := \Bbb Z$ so that $\mathcal C(N,-) = \operatorname{Hom}_{\mathbf{Ab}}(\Bbb Z,-)=\operatorname{id}$ so that the functor in question is really just $(-)\otimes_{\Bbb Z}M$, and this is representable iff $M$ is finitely generated and projective.
However, $M$ does not have to be finitely generated and projective in general. For instance, take $\mathcal C$ to be the one-object $\mathbf{Ab}$-category where with unique object $*$ and $\mathcal C(*,*)=\{0\}$ (note that this category is trivially copowered over $\mathbf{Ab}$). For any $\Bbb Z$-module $M$, the functor $\mathcal C(*,-)\otimes_{\Bbb Z}M:\mathcal C\to\mathbf{Ab}$ sends $*\mapsto\{0\}\otimes_{\Bbb Z}M=\{0\}$. Therefore, $\mathcal C(*,-)\otimes_{\Bbb Z}M=\mathcal C(*,-)$ is still representable.
In fact, we could have even taken $\mathcal C := \mathbf{Ab}$ and $N := \{0\}$ so that $\mathcal C(N,X)=\operatorname{Hom}_{\mathbf{Ab}}(\{0\},X) = \{0\}$. By the same argument as above, the composite $\operatorname{Hom}_{\mathbf{Ab}}(\{0\},-)\otimes_{\Bbb Z}M$ will also be corepresented by $\{0\}$ regardless of the choice of $\Bbb Z$-module $M$.