= Solution
Use <homogenization of Dirichlet boundary data>: put $v=u-\varphi$, using the supplied extension of $\varphi$ to the half-ball. Then $v=0$ on the flat face and $\Delta v=f-\Delta\varphi$. The <second-derivative estimate at a flat Dirichlet boundary> from part (a) gives
$$
\|v\|_{W^{2,2}(B_1^+)}
\leq C(n)\left(\|v\|_{W^{1,2}(B_2^+)}+\|f-\Delta\varphi\|_{L^2(B_2^+)}\right).
$$
The <triangle inequality> gives $\|v\|_{W^{1,2}}\leq\|u\|_{W^{1,2}}+\|\varphi\|_{W^{1,2}}$, while the <Cauchy-Schwarz inequality> bounds $\|\Delta\varphi\|_{L^2}$ by $\sqrt n\|D^2\varphi\|_{L^2}$. Finally $u=v+\varphi$, so restriction from the larger half-ball and another <triangle inequality> give
$$
\boxed{\|u\|_{W^{2,2}(B_1^+)}\leq C(n)\left(\|u\|_{W^{1,2}(B_2^+)}+\|f\|_{L^2(B_2^+)}+\|\varphi\|_{W^{2,2}(B_2^+)}\right).}
$$
\b[Nonzero boundary data contribute the second-order Sobolev norm of their extension.] No estimate for a separately constructed extension is needed, because $\varphi$ is already given on the half-ball.
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