If $G$ is a connected graph with least two cycles, it is true that for a base point vertex $v$, $\pi_1(G,v) \cong \mathbb{Z}^r$ for $r \geq 2$?
I think I have some intuition as to why this is the case, but I'm having trouble formalizing it. Suppose we're in the case where $G$ has two cycles. If one took a maximal spanning tree $T$ of $G$ and contracted it to $V$, one would be left with two circles which meet at $v$, giving rise to two non-trivial loops, and each contributing a generator (which won't have relations between them) to $\pi_1(G)$.
Will the converse hold as well? If $\pi_1(G) \cong \mathbb{Z}^r$, must $G$ have $r$ cycles?
No, the fundamental group of a connected graph with $r$ independent cycles ($r$ edges not in a spanning tree) is the free group on $r$ generators and is not Abelian for $r\ge2$. Its Abelianisation (quotient by its commutator subgroup) is $\Bbb Z^r$, so is determined up to isomorphism by $r$.
The free group consists of words in the letters $a_1,\ldots,a_r,a_1^{-1},\ldots,a_r^{-1}$ with the obvious relations $a_ja_j^{-1}=a_j^{-1}a_j=$ the identity. When $r\ge2$, $a_1a_2\ne a_2a_1$.