Let $M, M'$ be a profinite group. Let $M^*=\text{Hom}_{conti}(M,\Bbb{Q}/\Bbb{Z})$ be a Pontryagin dual of $M$. Let $f:M\to M'$ be a homomorphism of abelian group. Let $f^*: M'^*\to M^*$ be a map defined by $g\to f・g$.
Suppose $M'^*\cong M'$, then $f^*: M'\cong M'^*\to M^*$.
Does $\text{Im}(f)\cong \text{Ker}(f^*)$ hold? (In other words, is $M\stackrel{f}{\to}M' \stackrel{f^*}{\to}M^*$ exact ?)
I know $(\text{Im}(f))^* \cong \text{Im}(f^*)$ and $\text{Ker}(f^*) \cong (\text{Coker}(f))^*$, does the isomorphism $\text{Im}(f) \cong \text{Ker}(f^*)$ hold? I'm having difficulty relating the image of $f$ directly to the cokernel of $f$.
No. From the text you seem to be confusing cokernel and kernel; I am answering the question you wrote in symbols, which has kernel, and not the question you wrote in words, which has kernel.
Pick some isomorphism from $M$ to its dual. Now if $f$ is the identity map from $M$ to itself, then $f^*$ is also the identity map. So the image of $f$ is everything, and the kernel of $f^*$ is nothing, and therefore they are not the same under the chosen isomorphism.