But now, my secondary question is (in general). Are my current values correct?

Your current values seem correct.
According the datasheet of the PS2802, \$I_f\$ = 14 mA gives a \$V_f\$ = +/- 1.2V.
1.2V across 220 ohm yields a current of 5 mA. 14 mA + 5 mA = 19 mA = \$I_{total}\$.
19 mA through 1.2 kohm gives a 22.8V voltage drop.
And 22.8V + 1.2V makes 24V.

The other way around, when you don't know \$I_f\$ and \$V_f\$ yet (and you don't have LTspice or that specific component in LTspice) you want an expression of \$V_f\$ that gives a corresponding \$I_f\$ (or the other way around) to be used in the FORWARD CURRENT vs. FORWARD VOLTAGE figure given by the PS2802 datasheet.

So, let's start with

1. \$I_{total} = I_f + I_{R1}\$

The voltage across the LED of the PS2802 is the same across R1:

2. \$V_f = R1 \cdot I_{R1} = 220 \Omega \cdot I_{R1}\$

Rewriting gives

3. \$ I_{R1} = \frac{ V_f }{ 220 \Omega } \$

The current through R2 is the input voltage minus the forward voltage, divided by R2

4. \$I_{total} = \frac{ V_{in} - V_f}{R2} = \frac{24V}{1200 \Omega} - \frac{ V_f}{1200 \Omega} = 20 mA - \frac{ V_f}{1200 \Omega}\$

Substituting 4 in 1 gives

5. \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + I_{R1} \$

Substituting 3 in 5 gives

6. \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + \frac{ V_f }{ 220 \Omega }\$

Rewriting gives

7. \$ I_f = 20 mA - \frac{ V_f}{1200 \Omega} - \frac{ V_f }{ 220 \Omega }\$

8. \$ I_f = 20 mA - V_f \cdot \frac{71}{13200} \frac{ 1}{\Omega}\$

Now you can add (\$ I_f, V_f \$) points to the figure called FORWARD CURRENT vs. FORWARD VOLTAGE. (You cannot draw a linear line through these points, because the FORWARD CURRENT axis is logarithmic.)
Where these points intersect the (e.g.) 25°C curve, that point gives you the exact \$ V_f\$ and \$ I_f \$.

But now, my secondary question is (in general). Are my current values correct?

Your current values seem quite correct.
According the datasheet of the PS2802, \$I_f\$ = 13.5 mA gives a \$V_f\$ = +/- 1.2V.
1.2V across 220 ohm yields a current of 5.5 mA. 13.5 mA + 5.5 mA = 19 mA = \$I_{total}\$.
19 mA through 1.2 kohm gives a 22.8V voltage drop.
And 22.8V + 1.2V makes 24V.

The other way around, when you don't know \$I_f\$ and \$V_f\$ yet (and you don't have LTspice or that specific component in LTspice) you want an expression of \$V_f\$ that gives a corresponding \$I_f\$ (or the other way around) to be used in the FORWARD CURRENT vs. FORWARD VOLTAGE figure given by the PS2802 datasheet.

So, let's start with

1. \$I_{total} = I_f + I_{R1}\$

The voltage across the LED of the PS2802 is the same across R1:

2. \$V_f = R1 \cdot I_{R1} = 220 \Omega \cdot I_{R1}\$

Rewriting gives

3. \$ I_{R1} = \frac{ V_f }{ 220 \Omega } \$

The current through R2 is the input voltage minus the forward voltage, divided by R2

4. \$I_{total} = \frac{ V_{in} - V_f}{R2} = \frac{24V}{1200 \Omega} - \frac{ V_f}{1200 \Omega} = 20 mA - \frac{ V_f}{1200 \Omega}\$

Substituting 4 in 1 gives

5. \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + I_{R1} \$

Substituting 3 in 5 gives

6. \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + \frac{ V_f }{ 220 \Omega }\$

Rewriting gives

7. \$ I_f = 20 mA - \frac{ V_f}{1200 \Omega} - \frac{ V_f }{ 220 \Omega }\$

8. \$ I_f = 20 mA - V_f \cdot \frac{71}{13200} \frac{ 1}{\Omega}\$

Now you can add (\$ I_f, V_f \$) points to the figure called FORWARD CURRENT vs. FORWARD VOLTAGE. (You cannot draw a linear line through these points, because the FORWARD CURRENT axis is logarithmic.)
Where these points intersect the (e.g.) 25°C curve, that point gives you the exact \$ V_f\$ and \$ I_f \$.