1) implies 2):
Let $F$ be equicontinuous at $v_0$. Let $\varepsilon > 0$. By assumption we have that for $f \in F$ there is a $\delta_{v_0}$ such that $|v - v_0| < \delta_{v_0}$ implies $|f(v) - f(v_0)| < \varepsilon$. Now let $v_1$ be an arbitrary point.
We claim that if $|v - v_1| < \delta_{v_0}$ then $|f(v) - f(v_1)| < \varepsilon$.
Proof: Let $|v - v_1| < \delta_{v_0}$.
Then $|f(v) - f(v_1)| = |f(v - v_1)|$. Now translate $v_1$ to $v_0$ by subtracting $v_1 - v_0$:
$\begin{align} |f(v) - f(v_1)| = |f(v - v_1)| & = |f(v - (v_1 - v_0) + (v_1 - v_0) - v_1)| = \\ & = |f(v_0 - (v_1 - v)) -f(v_0)| \end{align}$
Now let $v\prime := v_0 - (v_1 -v)$. Then $|v^\prime - v_0| < \delta_{v_0}$ by assumption. Hence $|f(v) - f(v_1)| = |f(v^\prime) - f(v_0)|< \varepsilon $.
Hope this helps.