These two should be the standard examples for why Locally compact abelian groups are not an abelian categoty. The cokernel of any of these maps should is not a LCAG.
$$\Bbb Q^{\text{disc}} \hookrightarrow \Bbb R$$ $$\Bbb R^{\text{disc}} \hookrightarrow \Bbb R$$
But then, what is it? As a group, the first one seems to be the set of irrational numbers with the usual additive abelian structure. The cokernel of the second map looks like the zero object. But what is the topology?
Let's focus on the second example. The cokernel of $\mathbb{R}^\mathrm{disc} \to \mathbb{R}$ is indeed the trivial group $0$ (with the trivial topology), and the kernel is also trivial. But in an abelian category, any morphism with trivial kernel and cokernel is an isomorphism – which is clearly not the case for $\mathbb{R}^\mathrm{disc} \to \mathbb{R}$.
The cokernel of $\mathbb{Q}^\mathrm{disc} \to \mathbb{R}$ is more tricky. If you define locally compact groups to be Hausdorff, then the cokernel is trivial. (Indeed, if $X$ is a Hausdorff topological space, then any two continuous maps $\mathbb{R} \to X$ that agree on $\mathbb{Q}$ must be equal.) On the other hand, if you allow non-Hausdorff groups, the cokernel of $\mathbb{Q}^\mathrm{disc} \to \mathbb{R}$ is the group $\mathbb{R} / \mathbb{Q}$ with the (non-Hausdorff) trivial topology.