With resistors, the relationship between voltage across and current through them is well known, Ohms law, \$V=I \times R\$. With diodes (which includes LEDs, of course), this relationship is far more complex, dependent upon many factors (including temperature) which change from device to device, even between devices from the same manufacturing batch.
If you double the voltage across a resistance, the current through it will double. If you double the voltage across an LED (or any diode), the current through it won't double, it will increase many fold, perhaps by a factor of 10, 100 or more. This has the consequence that a diode will tend to keep a more-or-less constant voltage across itself even when the current through it varies wildly.
Here's a graph from the datasheet you linked to:
It shows the diode's current varying over two orders of magnitude (1mA to 100mA) while the voltage varies by only 1V (between 2.3V and 3.3V). The relationship is also non-linear, further complicating things. This diode (or any other) decides what voltage to have across it (in this case about 3V), and if you try to impose your own chosen voltage, the diode will fight back, effectively opposing any change in its voltage by conducting ridiculously large currents, or almost none at all.
In your circuit, disregarding the resistors, you have a battery (or some other source of voltage) with a fixed potential difference, and you are imposing that voltage directly across two diodes in series. Each diode will require about 3V, for a total of 6V. If your voltage source differs even slightly from 6V, either very little current will flow, and the LEDs will hardly glow at all, or a huge amount of current will flow, and the LEDs will glow too brightly, or burn out.
If you could guarantee that the two diodes were identical, and that each would share the total voltage equally between them (which is very unlikely), then you could approximate the current that would flow, using the graph above. With a voltage source of exactly 5.5V, then each diode would develop 2.75V, and from that graph we can see that current would be about 20mA.
However, if the voltage source were, say, to increase by half a volt, to 6V, then each diode would have 3V, and current would rise to 40mA. This is evidence enough that current is highly dependent on the voltage source, and even a small change of only 8% of that voltage can cause 100% change in current.
Armed with the exact equation of current vs. voltage for both LEDs (which is temperature dependent), and both voltage and temperature are well known to high precision, then you could employ Kirchhoff's voltage and current laws to obtain an expression for current, but I believe the exercise to be futile for the above reasons.
The resistors in your design are in parallel with the diodes, which (for reasons I won't go into here) would tend to balance the voltage across each diode, to mitigate any mismatch in their characteristics, but they are not solving these issues. The usual approach to this problem is to provide resistance in series with the diode (or diodes):
simulate this circuit – Schematic created using CircuitLab
Series resistors provide some flexibility, by developing whatever voltage the diodes don't want, whatever voltage remains after the diodes have taken whatever they take. They permit the diodes to choose their voltage, which can be seen if I measure diode voltage while varying battery voltage from 7V to 12V:
Each diode develops a reasonably constant 3V, regardless of battery voltage. The difference between battery voltage and diode voltage appears across the resistor.
The resistors also permit quite precise control of current. This graph of diode current vs. battery voltage shows a more linear relationship between current and voltage than the graph from the datasheet:
Clearly, with the resistance in place, a change in battery voltage will not cause the same enormous change in current as would occur if the battery was directly connected across the diode (or diode pair).
With your two parallel resistors in place, the mathematics to solve for current \$I\$ becomes very complex, since they add three more current paths, and three more variables to solve for. So my treatment will ignore them, and deal with only the diodes and a single resistance:
simulate this circuit
\$V_R\$ is the voltage across resistor R, and \$V_1\$ and \$V_2\$ are the voltages across the diodes. By Kirchhoff's Voltage Law (KVL), the sum of those voltages is the battery voltage:
$$ V_{BAT} = V_R + V_1 + V_2 $$
Kirchhoff's Current Law (KCL) tells us that the current in all elements here is the same, \$I\$, but we'll need equations relating that current to the voltages. For the resistor, this is trivial:
$$ V_R = I \times R $$
The diodes are more complex. We must a variant of the diode equation:
$$ V = ηV_T \ln \left(\frac{I}{I_S} + 1\right) $$
\$I_S\$ is the saturation current for the diode, and \$\eta\$ is the diode's "ideality". Neither of these values appear in the datasheet, so we're off to a bad start. \$V_T\$ is the thermal voltage, which will vary with temperature, but is about 25mV. If we assume the diodes are identical in every respect (again, not likely), then we can say:
$$ V_1 = V_2 = ηV_T \ln \left(\frac{I}{I_S} + 1\right) $$
Substituting into our KVL equation:
$$ V_{BAT} = IR + 2ηV_T \ln \left(\frac{I}{I_S} + 1\right) $$
Already you see that this will not be trivial to solve for \$I\$, even with \$R=0\$, and impossible in the absence of any diode parameters. If you can find those parameters, then you can indeed derive an expression for \$I\$, but you will find (if \$R=0\$, as is your case) that current varies wildly with even the slightest change in \$V_{BAT}\$. There will be no precision if \$V_{BAT}\$ isn't precisely known and constant.
However, we do know something about the diodes which will greatly simplify things, as long as we include \$R\$. We know that the voltage across the diodes is about 3V each, and remains near this value over a wide range of currents. By making this approximation, the KVL equation becomes:
$$
\begin{aligned}
V_{BAT} &= V_R + 3V + 3V \\ \\
&= IR + 6V
\end{aligned}
$$
I'm sure you'll agree, this is much simpler to solve. Choose your desired current \$I\$, and solve for \$R\$, to find the appropriate resistance.
For even better precision, use the diode's I-V curve (from the datasheet) to get a better approximation for the diode's voltage given your required current. For example, if you require 20mA of current, this corresponds to about 2.8V across each diode:
$$
\begin{aligned}
V_{BAT} &= V_R + 2\times2.8V \\ \\
&= IR + 5.6V
\end{aligned}
$$
