What is the difference between $(u \cdot \nabla)v$ and $u\cdot(\nabla v)$ when written in Einstein notation? I understand that they are different, but I'm not quite sure how. I've proven that $u \cdot (\nabla u)=\frac{1}{2}\nabla (u\cdot u)-u \times (\nabla \times u)$. I want to prove stuff with $(u \cdot \nabla)v$ but I don't know how to write it in Einstein notation.
Both $u$ and $v$ are vector fields.
On
If both $u$ and $v$ are vector fields, then $u\cdot(\nabla v) = (u\cdot\nabla)v$, since $$u\cdot(\nabla v) = u^{i}(\nabla_{i}v^{j}) = u^{i}\nabla_{i}v^{j}$$ $$(u\cdot\nabla)v = (u^{i}\nabla_{i})v^{j} = u^{i}\nabla_{i}v^{j} $$ i.e, there is no difference, both are different expressions for the same operation.
The formula you present is also valid with $(u\cdot\nabla)u$ $$u⋅(∇u) = (u\cdot\nabla)u = \frac{1}{2}∇(u⋅u)−u×(∇×u)$$ which, in index notation, is written as $$u^{i}\nabla_{i}u_{j} = \frac{1}{2}\nabla_{j}(u^{i}u_{i}) - {\epsilon_{j}}^{kl}u_{k}{\epsilon_{l}}^{mn}\nabla_{m}u_{n}$$
Where did you get the idea that $u\cdot(\nabla v)$ is different from $(u\cdot\nabla)v$?
It's important to note that in the first, $v$ is a vector function while in the second, $v$ is a scalar function. Try writing out each bracket first: $$\nabla v = \partial_i v\quad \text{so}\quad u \cdot (\nabla v) = u_i \partial_i v$$ while
$$u \cdot \nabla = u_i \partial_i \quad \text{so} \quad [(u \cdot \nabla) v]_j=u_i\partial_iv_j.$$