Am I applying KVL correctly to this very simple circuit?

You can greatly simplify the schematic. This will help you to create an independent means to work out the answer. So let's do that. No simulator, just yet.

^{simulate this circuit – Schematic created using CircuitLab}

In moving from the left schematic to the middle one, all I did was to swap the upper series pair and assign ground to one node. In moving from the middle schematic to the right one, all I did was get rid of the confusing clutter and label the node voltages, instead. (Ground is then assumed to be somewhere.)

From the right most schematic, find \$V_{_\text{X}}=\frac{V_3 R_1 R_2+V_2 R_1 R_3+V_1 R_2 R_3}{R_1 R_2+R_1 R_3+R_2 R_3}=600\:\text{mV}\$. To find the current magnitude in \$R_2\$ just compute \$\left|\frac{V_{_\text{X}}-V_2}{R_2}\right|=\left|\frac{600\:\text{mV}-\left(-3\:\text{V}\right)}{10\:\Omega}\right|=360\:\text{mA}\$.

So that makes sense.

Now let's do nodal (KCL), first. And here we'll use your original schematic on the left and ignore my simplifications. All the nodes are labeled there. Nice. I'll use your current names and directions, too.

$$\begin{align*} I_1+\frac{V_{_\text{X}}-V_{_\text{B}}}{R_2}+\frac{V_{_\text{X}}-V_{_\text{C}}}{R_3}&=0\:\text{A} \\\\ \frac{V_{_\text{A}}}{R_1} &= I_1 \\\\ \frac{V_{_\text{B}}-V_{_\text{X}}}{R_2} + I_2 &= 0\:\text{A} \\\\ \frac{V_{_\text{C}}-V_{_\text{X}}}{R_3} + I_3 &= 0\:\text{A} \end{align*}$$

That's KCL for all four nodes. In addition, you know that:

$$\begin{align*} V_{_\text{A}}+12\:\text{V}&=V_{_\text{X}} \\\\ V_{_\text{B}}&=-3\:\text{V} \\\\ V_{_\text{C}}&=-9\:\text{V} \end{align*}$$

You can either substitute into the earlier four equations in order to reduce unknown variables or else just keep all seven equations and seven unknowns (some of which are simply known already) and just solve.

Using Python and Sympy (both free) find:

var('i1 i2 i3 va vb vc vx r1 r2 r3') ans=solve([ Eq(i1+(vx-vb)/r2+(vx-vc)/r3,0), Eq(va/r1,i1), Eq((vb-vx)/r2+i2,0), Eq((vc-vx)/r3+i3,0), Eq(va+12,vx), Eq(vb,-3), Eq(vc,-9) ], [va,vb,vc,vx,i1,i2,i3]) for i9 in ans: i9,ans[i9].subs({r1:5,r2:10,r3:5}) (va, -57/5) (vb, -3) (vc, -9) (vx, 3/5) (i1, -57/25) (i2, 9/25) (i3, 48/25) 

So, \$V_{_\text{X}}=\frac35\:\text{V}=600\:\text{mV}\$ and \$I_2=\frac9{25}\:\text{A}=360\:\text{mA}\$. Just as before.

That's KCL.

For mesh (KVL), using \$I_1\$ for your top loop and \$I_2\$ for your bottom loop, as you show them to be in your last diagram:

$$\begin{align*} 0\:\text{V}-12\:\text{V}-R_1\cdot I_1-3\:\text{V}-R_2\cdot\left(I_1+I_2\right)&=0\:\text{V} \\\\ 0\:\text{V}-R_3\cdot I_2+9\:\text{V}-3\:\text{V}-R_2\cdot\left(I_1+I_2\right)&=0\:\text{V} \end{align*}$$

Python/Sympy:

ans=solve([ Eq(0-12-r1*i1-3-r2*(i1+i2),0), Eq(0-r3*i2+9-3-r2*(i1+i2),0)], [i1,i2]) for i9 in ans: i9,ans[i9].subs({r1:5,r2:10,r3:5}) (i1, -57/25) (i2, 48/25) 

The sum of these two currents is what's passing through \$R_2\$. Since both your arrows point from right to left, we get a negative value as the result of the sum which is just another way of saying that you got the direction wrong. The magnitude is still \$360\:\text{mA}\$.

So mesh (KVL) works.

I wasn't sure about which process you needed to improve, so I provided both just to be sure I got you covered.

It looks like you are mixing the "classic KVL/KCL method" with the mesh analysis.

Your circuit looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

Notice that I marked the polarity of voltage drops across the resistors. Using assumed currents directions.

Now we can write KVL for the first loop:

$$-12V - 5\Omega \cdot I_1 - 3V + 10\Omega \cdot I_2 = 0 $$

And KVL for the secend loop:

$$-5\Omega \cdot I_3 + 9V -3V+ 10\Omega \cdot I_2 = 0 $$

And one additional equation for KCL

$$I_1 + I_2 + I_3 = 0$$

And the solution is:

\$I_1 = -2.28A\$

\$I_2 = 0.36A\$

\$I_3 = 1.92A\$

This is what the classic KVL/KCL method looks like.