Well this $m$ is some element in the group $\mathbb{G}$ so there exists some $\alpha \in \mathbb{Z}_q$ such that $m=g^{\alpha}$, hence $m \cdot g^r = g^{\alpha+r}$ and notice that since $r$ is selected uniformly at random from $\mathbb{Z}_q$, then the distribution of $\alpha + r$ is exactly (not just computationally but exactly) uniform in $\mathbb{Z}_q$ as well.
Intuitively you can think of $r+\alpha$ as taking $\alpha$ which someone selected for you, and then adding to it a random number modulo $q$, and it is exactly uniform because for every possible result of $r+\alpha$ there is exactly a single $r$ that makes it that result, hence it is uniform.
As for the (a,b,r) - keep in mind, that the DDH assumption is trying to say something like: If you look at two random group elements $g^a, g^b$ and you have a third group element $h\in\mathbb{G}$, then you don't know if $h=g^{a\cdot b}$ or that $h$ is some completely random, unrelated group element $g^r$ for some random $r$.
The way you write this, is you look at two distributions where the $g^a, g^b$ are the same in both sides of the equality, but the third element is different (it is either $g^{a \cdot b}$ or $g^r$ and you say that under the DDH assumption, there doesn't exist an efficient probabilistic polynomial time turing machine that can distinguish between $g^{a \cdot b}$ and a random group element.
