Suppose $i$ and $j$ are indices which take values $1, \dots, m$, and for each $i$ and $j$, we have a number $a_{ij}$. Note that $(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}$. If we were to sum over all possible values of $(i, j)$ we would have
$$\sum_{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}}a_{ij}$$
which could also be written as
$$\sum_{i \in \{1, \dots, m\}}\sum_{j\in\{1, \dots, m\}}a_{ij}$$
or more commonly
$$\sum_{i=1}^m\sum_{j=1}^ma_{ij}.$$
Sometimes, we are not interested in all possible pairs of indices, but only those pairs which satisfy some condition. In the example you are looking at, the pairs of indices of interest are $(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}$ such that $i \leq j$. One way to denote the sum over all such pairs of indices is
$$\sum_{\substack{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}\\ i \leq j}}a_{ij}$$
but this is rather cumbersome. It would be much more helpful if we could write it as a double sum as above. To do this, note that we can list all suitable pairs of indices, by first fixing $i \in \{1, \dots, m\}$ and then allowing $j$ to vary from $i$ to $m$ (as these are the only possible values of $j$ with $i \leq j$). Doing this, we obtain the double sum
$$\sum_{i=1}^m\sum_{j=i}^ma_{ij}.$$
Note, we could also have fixed $j \in \{1, \dots, m\}$ and then allowed $i$ to vary from $1$ to $j$ (as these are the only possible values of $i$ with $i \leq j$). Doing this, we obtain an alternative double sum
$$\sum_{j=1}^m\sum_{i=1}^ja_{ij}.$$
The notation that you are asking about is yet another way to express the sum. That is,
$$\mathop{\sum\sum}_{i \leq j}a_{ij} = \sum_{\substack{(i, j) \in \{1, \dots, m\}\times\{1, \dots, m\}\\ i \leq j}}a_{ij} = \sum_{i=1}^m\sum_{j=i}^ma_{ij} = \sum_{j=1}^m\sum_{i=1}^ja_{ij}.$$
I think I've found a "correct" way of doing it.
Using the Kronecker delta, $\delta_{ij}$:
$$ \delta_{ij} =\left\{
\begin{array}{ll}
0 & i\not=j \\
1 & i=j \\
\end{array}
\right.
$$
Using that, we can get:
$$ \left(\frac{\partial u_1}{\partial x_1}\right)^2 + \left(\frac{\partial u_2}{\partial x_2}\right)^2 + \left(\frac{\partial u_3}{\partial x_3}\right)^2 = \frac{\partial u_j}{\partial x_k} \frac{\partial u_j}{\partial x_k} \delta_{jk} $$
This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).
Best Answer
$\sum_{i<j}$ sums over all the possible pairs $(i,j)$ for which $i<j$ holds. Similarly, $\sum_{i\neq j}$ sums over all the possible pairs $(i,j)$ for which $i\neq j$ holds.
For example, if $i$ and $j$ can take values in $\{1,2,3\}$, then
$$\sum_{i<j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,3},$$
whereas
$$\sum_{i\neq j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,1}+a_{2,3}+a_{3,1}+a_{3,2}.$$
If the summand is symmetric, i.e., $a_{i,j}=a_{j,i}$ holds for all $i$ and $j$, these two quantities are related by
$$\sum_{i\neq j}a_{i,j}=2\sum_{i<j}a_{i,j}.$$