Calculate zero sequence voltage and current in a HV network with one source and two feeders
&quot;Protective Relaying: Principles and Applications, Fourth Edition 4th Edition&quot; by J. Lewis Blackburn covers this well.
The source transformer you show (snip below) is not a path for zero sequence current (neutral ungrounded).

So, it looks like the only source of zero sequence current will be the wye-grounded capacitance on each line (probably referring to parasitic phase-ground  capacitance that is present on an ungrounded system). As such, this would be your symmetrical component model:

If your broken delta VT secondary (where you get U0) is loaded with a resistor (it likely would be in real life) then you reflect that to the primary as \$\frac{R}{3}*N^2\$ and it will go where I show \$Z_0\$ in the zero sequence. Otherwise just remove \$Z_0\$.
As for your directional element, the \$U_0\$ you show is what many protective relay engineers would call \$3V_0\$.  From symmetrical components, \$3V_0 = V_A + V_B + V_C\$.  In an ungrounded system like you have, \$3V_0\$ would have an amplitude \$\sqrt{3}V_{LL}\$.  You can see this from the phasor diagram that results when an A-phase to ground fault occurs on the ungrounded system.  The ground is now at A-phase bus so the B-phase and C-phase to ground voltages shift by 30° and lengthen by \$\sqrt{3}\$ compared to balanced positions.

If you do the math (equation above) or use head-to-tail on the phasors you will find \$3V_0\$ to be amplitude 3 (3 times higher than normal ph-ground voltage) and exactly 180° out-of-phase with the faulted phase (A in this case).  This can be your &quot;polarizing&quot; quantity for directional relaying.

A polarizing phasor is simply a phasor that has the same relative phase position for a fault down on Line 1 as it would for a fault down on Line 2.  This is what makes it suitable to compare against to determine fault direction. Note that in grounded systems (e.g. the vast majority of HV and EHV transmission) the protective relays are designed to use \$-3V_0\$ as the polarizing source to put the polarizing phasor and operate phasor (described below) in the same hemisphere on the phasor diagram - making the design easier to determine &quot;forward&quot; or &quot;reverse&quot; fault direction.  For zero-sequence voltage directional elements it is typical to have a characteristic like shown below (snip from Blackburn's book).  The \$V_{rs}\$ is the polarizing voltage (\$-3V_0\$) and the maximum torque line is located lagging that by 60°.  If the resulting &quot;operating&quot; quantity is on the &quot;operate zone&quot; side of the zero torque line (and is present above minimum volt-amp thresholds) then the directional element would call it a forward fault.

Now, all you have to do is calculate the zero sequence current (\$3I_0\$, also known as &quot;operate quantity&quot;) flowing to the right down Line 1 and to the right down Line 2 for your faults of interest.
This requires knowledge of symmetrical components and how to connect the three sequence networks for the fault of interest.  You also have to pay careful attention to polarity and recognize that your instrument VTs are subtractive polarity devices if standard.
Taking the symmetrical component networks and ignoring stuff that does not matter we can reduce the circuit to below where I have also connected (red lines) the 3 sequence networks for an A-ground fault on Line 1.

Now, further reducing things to &quot;what matters&quot; we get the following:

From this you can easily solve for the 3 quantities you care about, \$3I_0\$ in both Line 1 and in Line 2, and \$3V_0\$ at the bus.  From these 3 phasors you can see how to determine fault direction for each of your line relays.
UPDATE: To answer comment question.
Assuming the phase-ground capacitance is the same on both lines then we find:
$$I_{0_Line1} = \frac{V_F}{-jX_{C0}} = \frac{jV_F}{X_{C0}}$$
$$I_{0_Line2} = -I_{0_Line1}=\frac{-jV_F}{X_{C0}}$$
$$V_0 = I_{0_Line2}*-jX_{C0}=-V_F$$
Since this is an ungrounded system we will use \$3V_0\$ as our polarizing source instead of \$-3V_0\$.  This, with a maximum torque angle of -90°, allows our directional element to properly determine forward/reverse direction to the fault.  Below is a phasor diagram showing the relevant phasors for a fault down Line 1.

Note that the driving positive sequence voltage, \$V_F\$, is the pre-fault A-phase voltage which I set as 0° reference.