**PART 1:**

The element $u$ must have weight $-\alpha_2$, since $\mu = \lambda - \alpha_2.$ 

In $U(\mathfrak{n^-})$ there are only two linearly independent elements that have such weight (assuming PBW basis with respect to fixed order of generators based on positive roots): $y_{\alpha_2}$ and $y_{\alpha_1}y_{\alpha_3}.$ Hence the sought element $u$ is a linear combination of such
$$
u = a y_{\alpha_2}  + b y_{\alpha_1}y_{\alpha_3}.
$$

Since this has to be the image of the higest weight vector of $M(\mu)$ we must have $x_{\alpha_1} u = 0$ and $x_{\alpha_3} u = 0.$ Writing it out and using relations defining the Verma module and the Lie algebra, we end up with system of 2 linear equations for 2 unknowns. E.g. we have
$$
x_{\alpha_1} (ay_{\alpha_2}v_\lambda) = (a[x_{\alpha_1}, y_{\alpha_2}] + ay_{\alpha_2} x_{\alpha_1})v_\lambda
$$
where the first term on the right hand side is either zero, or some element of Cartan subalgebra acting on $v_\lambda$, and the second term is zero from the definition of the Verma module. 

**PART 2:**
I think you made a mistake in your calculations. For any $U(\mathfrak{g})$-homomorphism $\varphi$ we have $\varphi(u v) = u \varphi(v)$. Hence the composition going through $M(-2, 1, 1)$ is equal to $$
a_1y_{\alpha_3} \delta_{2_1}(v_{(-2, 1, 1)}) = a_1b_1 y_{\alpha_3} y_{\alpha_1}^2 v_{(0, -1, 1)}.
$$

The elements $y_{\alpha_1}$ and $y_{\alpha_3}$ do not commute, in fact $[y_{\alpha_1}, y_{\alpha_3}]$ should be a multiple of $y_{\alpha_2}.$ 

Similarly, the composition going through $M(-1,-1,2)$ equals
$$
a_2y_{\alpha_1} \delta_{2_1}(v_{(-1,-1,2)}) = a_2 b_2 y_{\alpha_1}(2y_{\alpha_1} y_{\alpha_3} + y_{\alpha_2})  v_{(0, -1, 1)}.
$$