Op-Amp Circuit Analysis Problem
Let's start with a few labels on the schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Assuming ideal opamps and a well-designed circuit operating within its limits, you know that \$V_b=V_1\$ and \$V_c=V_2\$. So it must be the case that \$i=\frac{V_1-V_2}{2\,R}\$. (I've changed the direction.) You also know that this exact same current must also be sourced from \$V_a\$ and sunk by \$V_d\$. So it follows that \$V_a=\frac12\left(3\,V_1-V_2\right)\$ and that \$V_d=\frac12\left(3\,V_2-V_1\right)\$. The difference is \$V_a-V_d=2\left(V_1-V_2\right)\$.

(There is another important symmetry about how \$V_a\$ and \$V_d\$ surround \$V_1\$ and \$V_2\$. But that's a different topic.)

Now, this is the place where I disagree with your solution (your second diagram), which appears to assume that the magnitude of the currents in the two resistors pointed at by blue arrows are the same. On its face, such a claim should seem false. The reasoning is simple enough. You know that \$V_e=V_f\$. But you also know that \$V_a\ne V_d\$ (unless \$V_1=V_2\$.) So how is it possible to sustain a claim that the current magnitudes (ignore direction) would be the same?

Instead, just find that \$V_e=V_f=\frac47\,V_d=\frac27\left(3\,V_2-V_1\right)\$ and then easily find that \$V_o=\frac83\left(V_2-V_1\right)\$. (You can derive the currents from there, if you like.)

I also liked the link that Nedd provided in a comment. The one here. But I chose to directly address your 2nd schematic, so I wrote differently here.

(To further clear your mind, just remember that the opamps are perfectly capable of sinking or sourcing current at their output nodes. There was no need to quantify those calculations here. But you could certainly work their contributions, too.)