The problem comes when trying to place two 'ideal' voltage sources in parallel. Ideal voltage sources have an internal resistance equal to 0 Ohms.
Let's think about it. These are two ideal voltage sources in parallel:

simulate this circuit – Schematic created using CircuitLab
The issue isn't obvious yet, but if \$V_1\neq V_2\$, as it is in practice, you have a contradiction.
In other words, if you put a voltmeter across \$V_1\$, is it going to give you the value of \$V_1\$? Or \$V_2\$? The ideal voltage sources enforce a value across the nodes they are placed. So in the previous case, you have two enforcing conditions for the voltage at the same node, which leads to a contradiction if the voltages are not the same.
The more general case looks like this:

simulate this circuit
Notice that if \$R_1=R_2=0\$, the above becomes the ideal case for voltage sources. Say, \$R_1\$ and \$R_2\$ are nonzero (probably \$\approx 0\$), then it doesn't matter what the values of \$V_1\$ and \$V_2\$ are—the current will flow from one source to the other for \$V_1\neq V_2\$ and we don't have a contradiction.
Now, back to your question, of "why the sources add up to the same". Imagine two perfectly matched sources (\$V_1= V_2\$ and \$R_1=R_2\$). Also notice that in the last circuit I have labeled two nodes: A and B. Then the voltage between A and B (what you'd think of 'parallel') is (after using KCL):
$$V_{AB}=V_1\dfrac{R_2}{R_1+R_2}+V_2\dfrac{R_1}{R_1+R_2} $$
Since we assume the sources are matched,(\$V_1= V_2\$ and \$R_1=R_2\$),
$$V_{AB}=V_1=V_2$$
If the sources aren't matched in some way, then \$V_{AB}\$ will be different. So when dealing with ideal voltage sources, you have to be careful as to not create a contradiction by placing two different enforcing conditions across the same nodes (similar argument can be said about ideal current sources). If you use the more realistic model, then the contradiction issue goes away but the parallel combination only adds up to the 'same' when the sources are matched.