Einstein Summation: $a_{kj}x_{j} + a_{ik} x_{i} = (a_{ik} + a_{ki})x_i$ If k is a constant, and $(1\le i \le n)$, and $(1\le j \le n)$.
Why does this Einstein Summation:
$$a_{kj}x_{j} + a_{ik} x_{i}$$
Equal this Einstein Summation:
$$a_{kj}x_{j} + a_{ik} x_{i} = (a_{ik} + a_{ki})x_i$$
(From Schaum's Outlines: Tensor Calculus, 2011, page 5-6, problem 1.10)
 A: It's just the distributive law. Write out an example with $n=2$ so only two values $x_1$ and $x_2$ and you will see that.
In the second term the name of the index is $i$ instead of $j$ but that's irrelevant. It could have been written
$$
a_{jk}x_j.
$$
Edit in response to comment.
$$
\begin{align}
\sum_{j=1}^2 a_{kj}x_j + \sum_{i=1}^2 a_{1k }x_i
 &  =  (a_{k1}x_1 + a_{k2}x_2) + (a_{1k}x_1 + a_{2k}x_2) \\
 &= (a_{k1}x_1 + a_{1k}x_1) + (a_{k2}x_2 + a_{2k}x_2)\\
& = (a_{k1} + a_{1k})x_1 + (a_{k2} + a_{2k})x_2 \\
 & = \sum_{z=1}^2 (a_{kz} + a_{zk})x_z \ .
\end{align}
$$
The $k$ is fixed throughout. It doesn't matter what indices you use for summation. I chose $z$ for the last sum.
A: $$=\sum_i^n a_{ki} x_i + \sum_j^n a_{jk} x_j$$
At this point we make a change of variables in the second summation by replacing j with i because its equivalent, and by doing so allows us to further simplify the expression....
$$=\sum_i^n a_{ki} x_i + \sum_i^n a_{ik} x_i$$
$$=\sum_i^n \bigg( a_{ki} x_i +  a_{ik} x_i \bigg)$$
$$=\sum_i^n x_i\bigg( a_{ki} +  a_{ik} \bigg)$$
$$=x_i ( a_{ki} +  a_{ik} )$$
