Your misunderstanding comes from assuming that the load side is always shorted to the transformer output. This is not the case. The voltage drop across the capacitor causes all of the diodes to turn off when the voltage falls below the peak so that the load is no longer shorted to the transformer and thus need not have the same voltage as it. Let me explain with some diagrams. Assume we have ideal diodes. Consider an input sine wave (on the transformer secondary terminals).
At $t=0$, the secondary voltage is zero, the capacitor voltage is zero and the load voltage is zero. All the diodes are shorted. No current is flowing. Everything is zero basically.
Now consider a time $t=0+dt$. The voltage drop across two of the diodes is now just barely positive, causing the diodes to short. The other two diodes are the opposite: they are open circuits now.
Just as you would expect, this would cause the capacitor and the load to both be at the voltage of the input. This continues until the input voltage reaches its positive peak at time $t=T$.
Now at time $t=T+dt$, the input voltage falls infinitesimally below its peak, and since the maximum input voltage is stored in the capacitor, the diodes which were previously shorted now have a slightly higher voltage at their negative terminal, causing them to become open circuits. This means the load is no longer shorted to the source and need not have the same voltage as it. The other diodes remain open circuits also for the same reason. So what results is this:
This means that the capacitor will now just discharge through the load. Now the input voltage is on its negative cycle. Once the absolute value of the input voltage just barely exceeds the capacitor's voltage, the other two diodes will short - recharging the capacitor until the input reaches its peak:
So in a summary, your analysis is correct - for only part of the input cycle:
For section A of the input cycle, two of the diodes are shorted like in the first image above. As you'd expect, the load voltage is thus the same as the rectified input voltage. However - and this is where you go wrong - in part B, all the diodes are open circuited, so only the capacitor determines the load voltage. In part C, the other two diodes are shorted, and the load voltage is again the same as the input rectified voltage.
Things are a little bit different in the non-ideal case, with forward diode drops and so on, but overall it's the same principle.
(Note: questions like these are better submitted to the Electronics Stack Exchange).