Electronic – measure (almost) no voltage between 230V phase and the body

1. I think the voltage drop in your top example is caused by the voltmeter's input impedance (probably around 10M) that slowly gets into range of the ohm-meter.
2. For range 20k and up it is again the voltmeter's input impedance issue. I think the 200Ω range is related to the diode measurement which requires a similar current source at a relatively high voltage. That leaves the 2kΩ range which is probably implemented in a cost effective way based on the current source for the 200Ω range.

Only with the circuit diagram the answer can be 100% sure.

Your multimeter will attempt to measure ohms by sending a known/set current through the attached resistor. This set current varies with the range your meter is in. However your multimeter has no ideal current source on board, but rather attempts to implement a current source from your battery voltage and a couple semiconductors, hence the open clamp voltage will never rise beyond the battery voltage.

Unsure why the voltage drops so much for the higher ranges, this will have to do with the way the current source is built. Notice that the 'high' voltage is not useful (forth column below) when you realize that the product of range times measurement current is much lower than the open clamp voltage (second column).

Also notice that the voltage measured in the lowest resistance range is identical to the voltage used for diode measurements for all three meters. For diode measurement you want a relatively high voltage to test the relatively high voltage drop across a diode. In that case you still use a constant current, but you are no longer interested in the resistance rather than the actual measured voltage. Useless to build two separate current sources for more or less the same current. On the other hand it is easier to build an accurate current source if you allow yourself a higher voltage drop across the current source and you don't need the voltage anyway (forth column).

Below are the results for my meters. For two out of three the input impedance of the voltmeter (10MΩ) was lower than the ohm-meter's range, so I skipped that value. The columns are as follows:

1. range
2. open clamp voltage
3. measurement current
4. maximum voltage required for measurement (range × current), notice how that voltage is reasonably constant!

DVM2000 (6V battery) \begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 3.25\text{V} &\Rightarrow& 785\text{µA}\\ 500Ω &\Rightarrow& 3.25\text{V} &\Rightarrow& 785\text{µA} &\Rightarrow& 500Ω × 785\text{µA} = 400\text{mV}\\ 5\text{kΩ} &\Rightarrow& 1.19\text{V} &\Rightarrow& 91.5\text{µA} &\Rightarrow& 5\text{kΩ} × 91.5\text{µA} = 460\text{mV}\\ 50\text{kΩ} &\Rightarrow& 1.18\text{V} ^{*)} &\Rightarrow& 11.5\text{µA} &\Rightarrow& 50\text{kΩ} × 11.5\text{µA} = 575\text{mV}\\ 500\text{kΩ} &\Rightarrow& 1.09\text{V} ^{*)} &\Rightarrow& 1.1\text{µA} &\Rightarrow& 500\text{kΩ} × 1.1\text{µA} = 550\text{mV}\\ 5\text{MΩ} &\Rightarrow& 614\text{mV} ^{*)} &\Rightarrow& 0.1\text{µA} \text{(last digit)}\\ 50\text{MΩ} &\Rightarrow& ? ^{*)} &\Rightarrow& ?\\ \end{array}

*) The open clamp voltage for ranges > 5kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 1.20V.

SBC811 (3V battery)

\begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 1.36\text{V} &\Rightarrow& 517\text{µA}\\ 200Ω &\Rightarrow& 1.36\text{V} &\Rightarrow& 517\text{µA} &\Rightarrow& 200Ω × 517\text{µA} = 103\text{mV}\\ 2\text{kΩ} &\Rightarrow& 645\text{mV} &\Rightarrow& 85.4\text{µA} &\Rightarrow& 2\text{kΩ} × 85.4\text{µA} = 171\text{mV}\\ 20\text{kΩ} &\Rightarrow& 645\text{mV} &\Rightarrow& 21.7\text{µA} &\Rightarrow& 20\text{kΩ} × 21.7\text{µA} = 434\text{mV}\\ 200\text{kΩ} &\Rightarrow& 637\text{mV} ^{*)} &\Rightarrow& 3.71\text{µA} &\Rightarrow& 200\text{kΩ} × 3.71\text{µA} = 742\text{mV}\\ 2\text{MΩ} &\Rightarrow& 563\text{mV} ^{*)}&\Rightarrow& 0.44\text{µA} &\Rightarrow& 2\text{MΩ} × 0.44\text{µA} = 880\text{mV}\\ 20\text{MΩ} &\Rightarrow& ? ^{*)} &\Rightarrow& 0.09\text{µA} \text{(last digit)}\\ \end{array}

*) The open clamp voltage for ranges > 2kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 645mV.

DT-830B (9V battery)

\begin{array}\\ \text{range} &\Rightarrow& \text{open clamp voltage} &\Rightarrow& \text{constant current} &\Rightarrow& \text{full scale voltage}\\ \hline\\ \text{diode} &\Rightarrow& 2.63\text{V} &\Rightarrow& 1123\text{µA} \\ 200Ω &\Rightarrow& 2.63\text{V} &\Rightarrow& 1123\text{µA} &\Rightarrow& 200Ω × 1123\text{µA} = 224\text{mV}\\ 2\text{kΩ} &\Rightarrow& 299\text{mV} &\Rightarrow& 70\text{µA} &\Rightarrow& 2\text{kΩ} × 70\text{µA} = 140\text{mV}\\ 20\text{kΩ} &\Rightarrow& 299\text{mV} &\Rightarrow& 23.0\text{µA} &\Rightarrow& 20\text{kΩ} × 23.0\text{µA} = 460\text{mV}\\ 200\text{kΩ} &\Rightarrow& 297\text{mV} ^{*)} &\Rightarrow& 2.95\text{µA} &\Rightarrow& 200\text{kΩ} × 2.95\text{µA} = 590\text{mV}\\ 2\text{MΩ} &\Rightarrow& 275\text{mV} ^{*)} &\Rightarrow& 0.35\text{µA} \text{(near scale low end)} &\Rightarrow& 2\text{MΩ} × 0.35\text{µA} = 700\text{mV}\\ \end{array}

*) The open clamp voltage for ranges > 20kΩ will probably be influenced by the 10MΩ input impedance of the voltmeter. They should probably all read 300mV.

If you don't connect the other AC wire between the capacitors, you will just have two caps in series being charged to around 325V through a full-wave rectifier bridge. But with the other AC wire connected between the caps, each cap gets the full 325V to itself during one half of the wave and will be effectively disconnected during the other half-wave.

To understand what's happening, draw the four diodes inside the bridge rectifier. Let's think of the midpoint between the caps as our ground reference - then the other AC wire will swing up to +325V and then down to -325V. When it swings up, current goes only through the upper capacitor and charges it, and the other cap just kind of dangles there thanks to the diodes preventing current flow. When the AC line swings down, only the lower cap gets charged. The output from both caps in series is double the input voltage peak.

Lower capacitance would only cause the output voltage to fall faster when there is a load connected. As for the wrong polarity, the only explanation is a wrong connection at some point. You can't always trust component markings, so check which way the diodes in the rectifier actually conduct!

For experimenting, I would suggest using the AC output from a small transformer, say 12V or so, and once you've seen it working correctly, only then connect it to 230V...