Intuitively
What follows is completely lacking in rigour, but you asked for intuitive, and here it is. I'm going to make these observations:
R1 forms a potential divider with R2. If the capacitor were not there, the potential at the junction of R1 and R2 would be:
$$ 10V \times \frac{R_1}{R_1+R_2} = 3\frac{1}{3}V = 3.33V $$
With C1 in place, this voltage will be C1's final, charged voltage.
C1's rate of charge is proportional to the current through it.
C1 is initially discharged, with 0V across it. Therefore R1 also has 0V across it, and consequently no current passes through R1, initially.
All current through R2 must therefore initally pass through C1, and this current will be the same in both cases (the full 10V across R2).
Initial rate of charge of C1 will be the same in both cases.
When R1 is present, the final "target" voltage is one third of the case when R1 is absent. Since the initial rate of change of voltage across C1 is the same in both cases, the capacitor will reach that target in one third of the time, with R1 in place.
That value of "one third" is key. It comes from the term \$\frac{R_1}{R_1+R_2}\$. I cannot think of an intuitive way of relating this to the combined parallel resistance R1 ∥ R2, except to point out:
$$ R_2 \times \frac{R_1}{R_1+R_2} = \frac{R_1 R_2}{R_1 + R_2} = R_1 \parallel R_2 $$
I confess I can't see how this helps, but at least you can see how the time constant is related to this parallel combination.
By Thevenin's Theorem
We can replace the entire network consisting of V1, R1 and R2 with its Thevenin equivalent circuit:
simulate this circuit – Schematic created using CircuitLab
\$R_{TH}\$ and \$V_{TH}\$ are calculated as follows:
$$
\begin{aligned}
V_{TH} &= V_1 \frac{R_1}{R_1 + R_2} \\ \\
R_{TH} &= R_1 \parallel R_2 = \frac{R_1R_2}{R_1 + R_2}
\end{aligned}
$$
These two circuits are functionally identical. If you enclosed them in a box, with only nodes A and B exposed, there's no way from the outside of distinguishing one from the other, because they behave identically. Any conditions you impose at nodes A and B of either box will yield exactly the same results.
So, if you connected a capacitor C1 between A and B, it would charge in the same way in both cases. Yet, in the second Thevenin equivalent circuit, clearly the time constant will be:
$$
\begin{aligned}
\tau &= R_{TH} \cdot C1 \\ \\
&= (R1 \parallel R2) \cdot C1
\end{aligned}
$$