To prove two functors are left and right adjoint

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In chapter 3 of the book the book Representations of $SL_2(\mathbb{F}_q)$ by Cédric Bonnafé: Harish-Chandra Induction, for two finite groups $\Gamma$, and $\Gamma'$, and $M$ a $(K[\Gamma],K[\Gamma'])$-bimodule where $K$ is an extension of the $l$-adic field containing $|\Gamma|$-th roots of unity and $|\Gamma'|$-th roots of unity (I dont think that the the base field is relevant apart from being char 0), define two functors $\mathcal F_M: K[\Gamma']-\text{mod} \to K[\Gamma]-\text{mod}$ seding $V'$ to $M\otimes V'$ and $^*\mathcal F_M: K[\Gamma]-\text{mod} \to K[\Gamma']-\text{mod}$ sending $V$ to $M^*\otimes V'$ where $M^* = \operatorname{Hom}(M,K)$ is also a $(K[\Gamma],K[\Gamma'])$-bimodule.

Since $K[\Gamma]$ and $K[\Gamma']$ are semisimple, the bimodule $M$ is projective both as a left $K[\Gamma]$-module and a right $K[\Gamma']$-module. The book now claims that the functors $\mathcal F_M,$ and $^*\mathcal F_M$ are left and right adjoint:

$$Hom_{K[\Gamma]}(V,\mathcal F_M V') \cong Hom_{K[\Gamma']}(^*\mathcal F_M V, V')$$ and
$$Hom_{K[\Gamma]}(\mathcal F_M V', V) \cong Hom_{K[\Gamma']}(V',^*\mathcal F_M V) $$

How does one prove this? This is what I tried:

I'm not sure if this is the right direction. If we let $f: V \to M \otimes V'$ be defined by $f(v) = \sum m_i \otimes v_i'$ we may define $\tilde f: M^*\otimes V \to V'$ as $\tilde f(\phi \otimes v) = \sum \phi(m_i) v_i'$. This is a well defined a homomorphism.

To prove that this is injective: If $\tilde f(\sum \phi \otimes v) = 0$ for all $\phi$ and $v$ then $\sum \phi(m_{i})v_{i}' = 0$. Assuming $f \neq 0$, pick $v$ such that $f(v) = \sum m_i \otimes v_i' \neq 0$ (all $m_i \neq 0$ linearly independent and all $v_i \neq 0$) and pick $\phi$ such that $\phi(m_i) = \delta_{i1}$, then $\tilde f(\phi\otimes v) \neq 0$, and injectivity follows.

I'm not sure about surjectivity. Seems like I'm missing some relation between $M$ and $M^*$.

It seems that the proof for the second isomorphism is similar, and here too, I have trouble proving surjectivity.