Forking but not dividing Definition.  A formula $\phi(x,a)$ divides over a set $B$ if there are $k<\mathbb{N}$ and  a sequence $(a_i)_{i<\omega}$ such that
(1) $\text{tp}(a/B)=\text{tp}(a_i/B)$, for all $i<\omega$;
(2)  $\{\phi(x,a_i)\}_{i<\omega}$ is $k$-inconsistent.

Definition. A formula $\phi(x,a)$ forks over a set $B$ if there are $n\in\mathbb{N}$ and formulas $\psi(x,b_1),\dots, \psi(x,b_n)$ such that
(1)  for each $i=1,\dots, n$, the formula $\psi_i(x,b_i)$ divides over $B$;
(2) $\phi(x,a)\models \bigvee_{i=1}^{n} \psi_i(x,b_i)$.

It is clear that dividing implies forking.
Question. Does forking always imply dividing?  If no, is there a formula that forks but does  not  divide?

No, forking does not always imply dividing. In simple theories dividing and forking are the same, but they are not the same in general. Look at the following example. 
Example. Let $\mathcal{L}=\{ R^{(3)} \}$ be a language which consists of a ternary relation. Consider the $\mathcal{L}$-structure $\mathcal{M}=\big(\mathbb{S}^1, R \big)$  where $\mathbb{S}^1$ is the  unit circle around the origin on the plane and $R(x,y,z)$ holds if and only if $y$ lies on the shorter arc between $x$ and $z$, ordered clock-wise,  including the endpoints. Now, let $a,b$ and $c$ be three equidistant points on $\mathbb{S}^1$. Then
$$\mathcal{M}\models \forall x\bigg( R(a,x,b)\vee R(b,x,c) \vee R(c,x,a)  \bigg)$$
Claim(1). Each of $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$.
Proof of Claim(1).
We will show, for instance, the formula  $R(a,x,b)$ 2-divides over $\emptyset$. Let $a, a_0,b_0, a_1,b_1,\dots, b$ be a sequence of consecutive points on $\mathbb{S}^1$.
Then   $(a_ib_i)_{i<\omega}$ is a sequence such that 
i) $\text{tp}(a_i,b_i)=\text{tp}(a,b)$ for each $i<\omega$;
ii)  $\big\{ R(a_i,x,b_i) \big\}_{i<\omega}$ is 2-inconsistent.
Therefore $R(a,x,b)$ 2-divides over $\emptyset$. Similarly we can prove that $R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$ as well.
Claim(2). The formula $x=x$ forks over $\emptyset$ but does not divide over $\emptyset$.
Proof of Claim(2). 
Since $x=x\models R(a,x,b)\vee R(b,x,c)\vee R(c,x,a)$ and  by Claim(1) each of  $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$, $x=x$ forks over $\emptyset$. But $x=x$ does not divide over $\emptyset$, because if $x=x$ divides over $\emptyset$, then there are $k<\omega$ and an $\emptyset$-indiscernible sequence $(a_i)_{i<\omega}$ such that $\big\{ x=x  \big\}_{i<\omega}$ is $k$-inconsistent which is a contradiction. 
Therefore the formula $x=x$ forks but does not divide.
