Why is $D_4/Z(D_4)$ "just like" $V$, Klein's 4-group? What makes the two similar? $D_4/Z(D_4)$ here is the quotient group of $D_4$ by its center.
There are a few computational details missing, but I will leave those for you to do.  After some computation, you will find that
$$ Z(D_4) = \{ e, r_2 \}, $$
where $r_2$ is the rotation of the square by $\frac{\pi}{2}$ radians.  Now $|Z(D_4)| = 2$, and so (recall the center of a group is always normal)
$$ |D_4 / Z(D_4)| = 4. $$
At this point, we may say that $D_4 / Z(D_4) \cong \mathbb{Z} / 4 \mathbb{Z}$ or $D_4 / Z(D_4) \cong V_4$ (the Klein $4$ group) because these are the only two groups of order $4$ up to isomorphism. Check that each element of $D_4 / Z(D_4)$ has order $2$, which implies $D_4 / Z(D_4) \cong V_4$.
Not only is D4/Z(D4) "like" the Klein 4 group, for all practical purposes it "is" the Klein 4 group, since it is isomorphic to the Klein 4 group. To see this fairly straightforwardly: 

We know that |Z(D4)| = 2, and so |D4/Z(D4)| = 8/2 = 4 by Lagrange's Theorem (We can consider D4/Z(D4), since the center is always normal in a group). Also notice that 4 = 2^2. 

Burnsides Theorem yields that any group of order p^2 where p is a prime is abelian. 

Thus, D4/Z(D4) is an abelian group of order 4. 

Now, by the fundamental theorem of finite abelain groups, D4/Z(D4) is either isomorphic to Z4 or Z2 X Z2. But Z4 is cyclic, and if D4/Z(D4) was isomorphic to Z4, then this would imply that D4/Z(D4) is also cyclic, which would imply D4 is abelian (by elementary theorem). This is clearly not true. 

Thus, D4/Z(D4) is isomorphic to Z2 X Z2, which is precisely the Klein 4 group.