EDIT#2: Here:
http://www.animations.physics.unsw.edu.au/jw/images/AC_files/resonancev(t).gif
is an image of the waveforms in this circuit.
You can't simply add the magnitudes of voltage drops across each element to get the maximum voltage. I am not entirely sure, but it looks like in your first equations you are simply writing the magnitudes of VC, VL and VR. You have to incorporate the phase of each. That's why your second method looks correct to me.
Think of it this way. If you are forming a right triangle that has sides, a, b, c such that:
$$ c^2 = a^2 + b^2 $$
If we look at sides a and b as vectors from the origin then we can't simply say that the sum of the magnitudes of a and b is c.
$$ c \ne a + b $$
We can say, however, that:
$$ \vec{c} = \vec{a} + \vec{b} $$
Use this analogy to think about your circuit. Inductors and capacitor have voltage phasors with angles:
$$ \vec{V_L} = IZ = j(I\omega L) = I\omega L\angle 90^\circ $$
$$ \vec{V_C} = IZ = \frac{I}{j\omega C} = -j\frac{I}{\omega C} = \frac{I}{\omega C}\angle{-}90^\circ $$
$$ \vec{V_R} = IZ = IR $$
Then we can get the vector sum, and take its magnitude:
$$ \vec{V_{total}} = \vec{V_L} + \vec{V_C} + \vec{V_R} $$
$$ \vec{\left\lvert V_{total} \right\lvert} = \left\lvert \vec{V_L} + \vec{V_C} + \vec{V_R} \right\lvert $$
In short, the angles are important and because part of the phasor sum.
EDIT: added vector notation for clarity with phasors
The defining equation for a capacitor is $Q=CV_{\rm C}$ and when that equation is differentiated with respect to time one gets $\dfrac{dQ}{dt} = I = C\dfrac{dV_{\rm C}}{dt}$
So the current is proportional to the rate of change of voltage across the capacitor
Applying a sinusoidal voltage to a capacitor results it the following current and voltage graphs.
Notice that the current is determined by the gradient of the voltage against time graph. being a maximum at time $a$ and zero at times $b$and $d$.
Whatever the current is doing the voltage does a quarter of a period (equivalent to $90^\circ$) later.
So the current is a maximum at time $a$ and the voltage is a maximum at a later time $b$.
We say that current leads the voltage across a capacitor by $90^\circ$.
In the graph $V_{\rm C}(t)=V_{\rm max} \sin \omega t$ and so the current is $I(t) =\omega CV_{\rm max} \cos\omega t$ with a peak current $I_{\rm max}=\omega CV_{\rm max}$.
When you add a series resistor to the circuit the current is the same in all parts of the circuit.
The voltage across the capacitor still lags the current by $90^\circ$ and the voltage across the resistor will be in phase with the current.
The (applied) voltage across both components will lag the current through the circuit at some value between $0^\circ$ and $90^\circ$ depending on the values of the capacitance of the capacitor, the resistance of the resistor and the frequency of the applied voltage.
Here I have considered what are called steady state conditions and so there are no transients which would be characterised by an exponential function and a time constant.
The difference for an inductor is that the defining equation is $V_{\rm L} = L \dfrac{dI}{dt}$ and the voltage across an inductor leads the current by $90^\circ$.
Update as a result of a comment
I think that what you are asking about is the transient behaviour which occurs when you first connect a capacitor to the voltage source. If by chance you make this connection to an uncharged capacitor when the voltage of the supply is zero then there is no transient and the circuit currents and voltages are as per the graph shown above. If on the other that is not so you will have a combination of the transient (the exponential function you have described) and the steady state. After about 10 time constants (10CR) the transients would have decayed away and all that is left is steady state
Now with an "ideal" circuit with no resistance the time constant is zero and the circuit settles down to steady state behaviour "instantly". However with a finite resistance in the circuit then there will be a transient behaviour which you tend to to see because it decays away.
I can show you this idea of a transient in action by using the "Circuit Sandbox" which is available in the edX Circuits and Electronics course, a course I thoroughly recommend even if it just to be able to used the circuit simulator.
Here is the result of a simulation where there is dc voltage of 1 V across a capacitor and after one second a sinusoidal voltage of peak value 1 V and frequency 1 Hz is applied across a resistor and a capacitor connected in series.
The graph is voltage across the capacitor in volts against time in seconds.
You can see very clearly the transient behaviour (the exponential decay) and then the steady state behaviour.
The 10RC s just a rule of thumb where $e^{-10} \approx 4.5 \times 10^{-5}$ and the decay has effective finished.
Others use 5RC which corresponds to a decrease of $e^{-5} \approx 6.7 \times 10^{-3}$.
Update 2
Here is the supply voltage shown in red and the voltage across the capacitor shown in cyan.
The supply voltage and voltage across the capacitor start at $+1 \, \rm V$ and then a $\pm 1 \, \rm V$ sinusoidal voltage is added after 1 second.
It clearly shows the $90^\circ$ phase shift.
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Best Answer
Well the condition will be different for each component. Resonance doesn't matter. The current must obviously be identical in all three series elements. The Voltage across the resistance will be in phase with that current, and the Voltage across the capacitance must lag the current by 90 degrees, while the Voltage across the inductance must lead the current by 90 degrees. This makes the inductance and capacitance Voltages 180 degrees out of phase. At resonance they will be equal and opposite, so they cancel leaving just the whole source Voltage across the resistance.