You got off to great start, ignoring D1 for the time being, and focussing on that chain of R1, D2 and R2.
Before I address where your calculation failed, let me do a more formal analysis, to establish the truth:
KVL says that all the voltages across R1, D2 and R2 must add up to the total voltage between the two ends of the chain, 10V:
$$ V_{R1} + V_{D2} + V_{R2} = 10 $$
KCL tells us that the current down the entire chain (as long as we ignore current branching away via D1) is the same in every element, which I can write as:
$$ I_{R1} = I_{D2} = I_{R2} = I $$
The last thing we do is apply Ohm's law to glue everything together. We can relate \$I\$ to \$V_{R1}\$ and \$V_{R2}\$ like this:
$$ V_{R1} = I \times R_1 $$
$$ V_{R2} = I \times R_2 $$
Substitute these last two into the first equation, and also plug in \$V_{D2}=0.7V\$:
$$ I \times R_1 + 0.7 + I \times R_2 = 10 $$
That's enough to find \$I\$, which I will leave for you to do. Once you know \$I\$, it's trivial to find \$V_{R1}\$ and \$V_{R2}\$ from the Ohm's law equations above.
The value of \$v_o\$ is also found with an application of KVL, although you might not have thought of it in that way. The potential at the top of R2 must be higher than the potential at the bottom of R2, by an amount \$V_{R2}\$:
$$
\begin{aligned}
v_o &= 0V + V_{R2} \\ \\
&= V_{R2}
\end{aligned}
$$
Maybe you can already see where you went wrong. Your initial calculation for current through R1 was incorrect, because you failed to account for all the elements in the chain.
Intuitively, the voltage remaining after the diode has "removed" or "taken up" its share of 0.7V is \$10V - 0.7V = 9.3V\$. However this "remainder" is shared by both R1 and R2.
You seem to have assumed that this voltage would appear across only R1, but actually it is shared by all remaining elements.
From that perspective, current flowing must be due to 9.3V across R1 and R2:
$$ I = \frac{9.3V}{10k\Omega + 10k\Omega} $$
When you know \$v_0\$ with D1 absent, you can start thinking about the influence of \$v_i\$ and D1.
I'll redraw the circuit with a named node x at the junction of the two diodes:
simulate this circuit – Schematic created using CircuitLab
The potential \$v_x\$ is 0.7V higher than \$v_o\$:
$$ v_x = v_o + 0.7 $$
Let me define a value \$v_{xu}\$ to be the potential at x with D1 absent. Since \$v_x\$ will be varying, I prefer to have a different variable referring to its "unloaded" value, a constant with value \$v_o + 0.7\$ when \$v_i\$ and D1 are disregarded.
It is only possible for \$v_i\$ to influence \$v_x\$ (and \$v_o\$) when D1 is conducting, forward biased. It's pretty clear that for \$v_i > v_{xu}\$ it can't possibly have any influence, since D1 is reverse biased. That portion of the graph must be flat, since \$v_o\$ is unchanging.
It's also pretty easy to see that for D1 to be conducting (with 0.7V across it), \$v_i\$ must be lower than 0.7V below \$v_{xu}\$. In other words, when \$v_i\$ drops to \$v_{xu} - 0.7V\$ the diode becomes fully conductive.
As \$v_i\$ falls further, it drags \$v_x\$ with it, such that \$v_x = v_i + 0.7V\$. This is the relationship that determines the section of graph \$v_i < (v_{xu}-0.7)\$.
Don't forget, you're plotting \$v_o\$ vs. \$v_i\$, not \$v_x\$.
