Your axiom .2 is $\mathrm{G1}$ from *Hughes&Cresswell*, so I will use this name. They (and also Wikipedia) call the corresponding frames *convergent* (another name for *directed*). Also, I am going to rewrite axiom $\mathrm{C}$ as
$(\Diamond p \land \Box(\Box p \rightarrow p) \land \Box(p \rightarrow \Diamond p)) \rightarrow \Box p$.

*$\mathrm{C}$ is not derivable in $\mathbf{KD4G1}$.*

If it were, then it would also be derivable in $\mathbf{S4.2}$, which extends $\mathbf{KD4G1}$.
But $\mathrm{C}$ plus $\mathbf{S4.2}$ reflexivity immediately implies
$\Diamond p \rightarrow \Box p$,
then $p \rightarrow \Box p$, therefore $p \leftrightarrow \Box p$
and $\mathbf{S4.2}$ would collapse to $\mathbf{Triv}$,
which is known not to be the case.

Note that the frame condition you give for C is incorrect,
since it implies seriality when $y_0=z_0=y$,
while C is valid on the non-serial frame $F=(\{x,y\},\{(x,y)\})$.
Assume we fix this by weakening the definition of ancestral convergence to $y_0\not=z_0$.
Then we still get only a necessary, but not sufficient, condition for a C-frame.
Indeed, the frame
$F'=(\{x,y,z\},\{(x,y),(x,z),(y,y),(y,z),(z,z)\})$ is serial, transitive and convergent, therefore ancestrally convergent,
yet $\mathrm{C}$ fails to validate at $x$ for valuation $V(p)=\{y\}$.
And $F'$ validates $\mathrm{G1}$, which is another proof that $\mathrm{C}$ is not derivable in $\mathbf{KD4G1}$.

*$\mathrm{G1}$ is derivable in $\mathbf{KD4C}$.*

Here is an outline of the proof:

a) In $\mathbf{KD4}$, derive
$(\Box p \land \Box\Box p) \leftrightarrow \Box p$ and
$\Box p \rightarrow \Box\Diamond p$ and
$\Box\Diamond p \rightarrow \Diamond p$ and
$\Box p \rightarrow \Diamond\Box p$ and
$\Diamond (p \land \Box p) \leftrightarrow \Diamond\Box p$,
all quite straightforward.

b) Substitute $p\land\Box p/p$ in $\mathrm{C}$, then use all a) and simplify it to
$(\Diamond\Box p \land \Box(\Box p \rightarrow p)) \rightarrow \Box p$.

c) Substitute $\Diamond p/p$ in the result of b), then apply 3-rd a) and simplify it to
$\Diamond\Box\Diamond p \rightarrow \Box\Diamond p$.

d) Apply $\Diamond$-monotony to 2-nd a) and derive
$\Diamond\Box p \rightarrow \Diamond\Box\Diamond p$.

e) $\mathrm{G1}$, i.e. $\Diamond\Box p \rightarrow \Box\Diamond p$, follows from c) and d).

3more comments