I want to prove that $\dim_{\mathbb{Z}_2}(H_k(M;\mathbb{Z}_2))=\dim_{\mathbb{Z}_2}(H_{n-k}(M,\partial M;\mathbb{Z}_2))$.
This seems like a trivial result from Poincaré-Duality, so I tried \begin{align} H_{n-k}(M,\partial M;\mathbb{Z}_2)=&H^k(M;\mathbb{Z}_2)\\ =&Hom(H_k(M;\mathbb{Z}_2))+Ext(H_{k-1}(M;\mathbb{Z}_2);\mathbb{Z}_2) \end{align} by using PD and universal coefficient theorem. Now I am unsure of how to get rid of the $Ext$-functor. I know the result would fit if $H_{k-1}$ would be free, but since that is not given I am clueless and I am generally unsure how to approach the $\mathbb{Z}_2$-coefficients.
Specifically the universal coefficient theorem says there is a short exact sequence
$$ 0\to Ext_R^1(H_{n-1}(X; R), G) \to H^n(X;G) \to Hom_R(H_n(X;R), G) \to 0 $$
for any principal ideal domain $R$ and $R$-module $G$. In your case both $R$ and $G$ are $\mathbb{Z}/2$, but $\mathbb{Z}/2$ is free as a module over itself so the $Ext$ term vanishes.
More to the point, whenever $\mathbb{F}$ is a field the term "module" means "vector space" so ALL modules are free, so the $Ext_{\mathbb{F}}$ term ALWAYS vanishes and $$H^n(X;G) \cong Hom_{\mathbb{F}}(H_n(X;\mathbb{F}), G) = Lin(H_n(X;\mathbb{F}), G)$$ for any $\mathbb{F}$-vector space $G$.
Edit: in light of some comments I should clarify that in the context of $R$-modules, "free" means "has a basis": https://en.wikipedia.org/wiki/Free_module. $\mathbb{Z}/2$ can be viewed as a $1$-dimensional vector space over itself.