You cannot get the exact phase-ground voltages because the delta connection removes the zero sequence component. For example, below I show the symmetrical component equation for a & b phase voltages in an abc rotation system,
$$V_{an}=V_0 + V_1 + V_2 $$
$$V_{bn}=V_0 + \alpha^2V_1 + \alpha V_2 $$
So, the a-b voltage will be,
$$V_{ab}=V_{an}-V_{bn}$$
$$V_{ab}=(V_0 + V_1 + V_2)-(V_0 + \alpha^2V_1 + \alpha V_2)$$
$$V_{ab}=V_1(1-\alpha^2) + V_2(1-\alpha)$$
So, you can see that the zero sequence is removed by the delta connection, never to return.
If you want to assume the system is perfectly balanced (with no zero-sequence or negative-sequence) then you can just shift by 30° and scale by \$\frac{1}{\sqrt3}\$ as per your phasor diagram.
Also, you could change the transformer to a Ynyn (wye-grounded/wye-grounded) and get exactly what you need.
Doing the Clarke transformation would buy you nothing with your present configuration.
UPDATE: Answering comment question about calculating phase-ground voltages from phase-phase voltages ignoring zero-sequence.
From C.L. Fortesue's definition of his symmetrical components we know (where \$\alpha=1∠120°\$),
$$V_{ab} = (V_0+V_1+V_2)-(V_0-\alpha^2V_1-\alpha V_2)$$
$$V_{bc} = (V_0+\alpha^2V_1+\alpha V_2)-(V_0-\alpha V_1-\alpha^2V_2)$$
$$V_{ca} = (V_0+\alpha V_1+\alpha^2V_2)-(V_0-V_1-V_2)$$
which reduces to,
$$V_{ab} = V_1(1-\alpha^2)+V_2(1-\alpha)$$
$$V_{bc} = V_1(\alpha^2-\alpha)+V_2(\alpha-\alpha^2)$$
$$V_{ca} = V_1(\alpha-1)+V_2(\alpha^2-1)$$
Now taking the differences of the above we can create a set of equations to solve,
$$V_{ab}-V_{bc} = V_1(1-2\alpha^2-\alpha)+V_2(1-2\alpha+\alpha^2)$$
$$V_{bc}-V_{ca} = V_1(\alpha^2-2\alpha+1)+V_2(\alpha-2\alpha^2+1)$$
which reduces to,
$$V_{ab}-V_{bc} = V_1(3∠60°)+V_2(3∠{-60°})$$
$$V_{bc}-V_{ca} = V_1(3∠{-60°})+V_2(3∠60°)$$
and in matrix form,
$$\begin{pmatrix}3∠60° & 3∠{-60°} \\\ 3∠{-60°} & 3∠60°\end{pmatrix} \begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} V_{ab}-V_{bc} \\\ V_{bc}-V_{ca} \end{pmatrix}$$
which is easy to solve and subsequently compute the phase-ground phasors.
Example (using per unit and assumes abc rotation system): Let \$V_a=1∠0° pu\$, \$V_b=1.12∠-130° pu\$, and \$V_c=0.95∠122° pu\$
$$V_{ab}=1.924∠26.54° pu$$
$$V_{bc}=1.680∠{-97.46°} pu$$
$$V_{ca}=1.706∠151.81° pu$$
So,
$$V_{ab}-V_{bc}=3.18∠52.5° pu$$
$$V_{bc}-V_{ca}=2.78∠{-62.5°} pu$$
$$\begin{pmatrix}3∠60° & 3∠{-60°} \\\ 3∠{-60°} & 3∠60°\end{pmatrix} \begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} 3.18∠52.5° \\\ 2.78∠-62.5° \end{pmatrix}$$
with solution,
$$\begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} 1.019∠{-3°} \\\ 0.091∠51.3° \end{pmatrix}$$
Plugging these values into the fundamental symmetrical component equations (with assumption \$V_0=0\$) we have,
$$V_{an}=V_0 + V_1 + V_2=1.075∠0.94° $$
$$V_{bn}=V_0 + \alpha^2V_1 + \alpha V_2=1.06∠{-127.5°} $$
$$V_{cn}=V_0 + \alpha V_1 + \alpha^2V_2=0.928∠117.56° $$
Now, as a check, since we know \$V_0=(1∠0°+1.12∠-130°+0.95∠122°)/3 = 0.076∠-166.8° pu\$ we can add it to each of the above 3 values to get the actual original phase-neutral voltages. This shows us the error we could expect if zero-sequence voltage is not negligible (depends on what you are trying to accomplish).
$$V_{an}=V_0 + 1.075∠0.94° = 1∠0°$$
$$V_{bn}=V_0 + 1.06∠{-127.5°} = 1.12∠-130°$$
$$V_{cn}=V_0 + 0.928∠117.56° = 0.95∠122°$$
Finally, if what you are really interested in are the phase-ground voltages on the high-voltage side (left side in your first drawing) then you simply need to translate \$V_1\$ and \$V_2\$ across the transformer first. The transformer you drew has low-side leading by 30° (\$V_{an}\$ in phase with \$V_{AB}\$). So, you would multiply \$V_1\$ by the voltage ratio (HV ph-ph divided by LV ph-ph) and shift it's phase angle by \$-30°\$. You would multiply \$V_2\$ by the voltage ratio and shift it's angle by \$+30°\$. From those 2 quantities you can now calculate the high-voltage side phase-ground voltages (of course, assuming \$V_0=0\$).