You have a homomorphism from the free abelian group of rank $3$, generated by $x$, $y$, and $z$, onto $V$, with kernel given by the equations: $3x+2y+8z=0$, $2x+4z=0$. Finding the kernel is the same as finding the integral kernel of the matrix $\left(\begin{array}{ccc} 3 & 2 & 8\\ 2 & 0 & 4 \end{array}\right)$ which can be found using Gaussian elimination or Smith Normal form. $\begin{align*} \left(\begin{array}{ccc} 3 & 2& 8\\ 2 & 0 & 4 \end{array}\right) &\to \left(\begin{array}{ccc} 1 & 2 & 4\\ 2 & 0 & 4 \end{array}\right) &&\to \left(\begin{array}{rrr} 1 & 2 & 4\\ 0 & -4 & -4 \end{array}\right)\\ &\to\left(\begin{array}{rrr} 1 & 2& 4\\ 0 & 4 & 4 \end{array}\right). \end{align*}$ This tells you that $x+2y+4z =0$ and $4y+4z=0$. This in turn says that you can get $x$ from $y$ and $z$, that $y+z$ has order $4$, So you can generate the group with $y+z$ and $z$, subject only to the condition that $y+z$ has order $4$. This gives you the cyclic summand of order $4$ (for $\langle y+z\rangle$), and an infinite cyclic group (for $z$).