How to calculate the total error from an arbitrary hidden layer in a neural network back propogation?
The short answer is yes, $\frac{\partial E_{total}}{\partial w_1} = \frac{\partial E_{j_1}}{\partial w_1} + \frac{\partial E_{j_2}}{\partial w_1}$.

The long answer. The key formula is the chain rule, as D.W. mentioned:

$\frac{\partial E_{total}}{\partial w_1} = \frac{\partial E_{total}}{\partial out_{h_1}} \cdot \frac{\partial out_{h_1}}{\partial net_{h_1}} \cdot \frac{\partial net_{h_1}}{\partial w_1}$

What's good about it is that all three components on the left are local information. No matter what the next or previous layers are, 

$\frac{\partial out_{h_1}}{\partial net_{h_1}} = out_{h_1} (1 - out_{h_1})$ 

$\frac{\partial net_{h_1}}{\partial w_1} = i_1$

Hence, the calculation is the current node depends only on forward messages from direct neighbors to the left and backward messages from direct neighbors to the right:

$out_{h_1}$ and $i_1$ are known from the forward pass, and $\frac{\partial E_{total}}{\partial out_{h_1}}$ is the total backward message.

In the architecture from the post, the node $h_1$ has two direct neighbors to the right: $o_1$ and $o_2$, and that explains the sum $\frac{\partial E_{total}}{\partial out_{h_1}} = \frac{\partial E_{o_1}}{\partial out_{h_1}} + \frac{\partial E_{o_2}}{\partial out_{h_1}}$. In your example, the neighbors are $j_1$ and $j_2$, so it will be the sum  $\frac{\partial E_{total}}{\partial out_{h_1}} = \frac{\partial E_{j_1}}{\partial out_{h_1}} + \frac{\partial E_{j_2}}{\partial out_{h_1}}$. If $h_1$ has even more connections, all of them will pass the backward message and they will be added up.