In an RC circuit, how do you explain the fast rise of voltage...

 I answered this SE EE question yesterday.

Since you want to intuitively understand the circuit, I will suggest a few intuitive approaches.

"Pulling" resistors

When there is a resistor in between source and capacitor, how do we know which effect predominates? Pull up of source or slow rise of capacitor voltage?

In fact, you came up with the idea yourself by imagining the circuit as two voltage sources (V1 and VC) "pulling" the common midpoint through two resis

tors - the "pull-up" R1 and "pull-down" R2. Each of the voltage sources tries to set its voltage at midpoint. Since VC does not "move" when V1 "jumps", the midpoint also "jumps" but weaker.

"Shifted" voltage divider

As other said, R1 and R2 constitute a voltage divider. This is an ordinary voltage divider only in the first moment when V1 "jumps". After that, this is a "shifted" voltage divider since the lower end of R2 is not grounded but is slowly "moving".

Voltage summer

Actually, the network of two resistors in series act as a voltage summer with weighted inputs (it can be easily analyzed by the superposition principle). So the V1 "jump" appears at the summer's output (the midpoint).

Voltage diagram

But the most interesting representation would be to "open" the resistors to "see" the voltage at each point of the resistive film. I started doing it (real and mental, using my imagination as a "simulator") from 1991 until now. I have attached below my authentic recordings from that time, unfortunately in Bulgarian. But I think the illustrations are understandable (if there is interest, I can do a full translation).

The idea was simple - to represent each local voltage by a segment of proportional length; thus I got a "voltage diagram" as a set of bars.

For simplicity, we can depict only the envelope of the diagram (as a triangle).

It is interesting that, 30 years ago, I asked myself the same question about the RC integrating circuit and visualized it using such a voltage diagram. So, when the input voltage V1 (E in the picture) "jumps", all local voltages along the resistive film instantly and proportionally rise from right to left.

Then, when VC (Uc) slowly rises, all local voltages slowly and proportionally rise from left to right.

Finally, the capacitor is charged and all local voltages along the resistive film are equal to V1 (E).

When V1 (E) becomes zero again, the capacitor becomes a "voltage source" and begins discharging through V1 to ground.

A more special case is when V1 (E) alternatively varies and the capacitor stays uncharged.