Which way of handling inequality constraints do I use?

Step 1 — weak duality (always true, no convexity needed). Define the Lagrangian L(x,\lambda,\nu) = f_0(x) + \sum_i \lambda_i f_i(x) + \nu^\top(Ax-b) and the dual function g(\lambda,\nu) = \inf_x L(x,\lambda,\nu). For any \lambda\ge0 and any primal-feasible x, every term \lambda_i f_i(x)\le0 and \nu^\top(Ax-b)=0, so L(x,\lambda,\nu)\le f_0(x); taking the infimum over x preserves this, so g(\lambda,\nu)\le f_0(x) for every feasible x, hence g(\lambda,\nu)\le p^* for every dual-feasible (\lambda,\nu) — this is weak-duality-lp's general (nonlinear) counterpart, and gives d^*=\sup g \le p^* always.

Step 2 — Slater's condition closes the gap. Consider the set in \mathbb{R}^m\times\mathbb{R} \mathcal A = \{(u,t) : \exists x,\ Ax=b,\ f_i(x)\le u_i\ \forall i,\ f_0(x)\le t\}. Convexity of each f_i makes \mathcal A convex: given (u,t),(u',t')\in \mathcal A witnessed by x,x', the point x''=\theta x+(1-\theta)x' satisfies Ax''=b and, by convexity of each f_i, f_i(x'')\le \theta f_i(x)+(1-\theta)f_i(x') \le \theta u_i+(1-\theta)u_i', so (\theta u+(1-\theta)u',\, \theta t+(1-\theta)t')\in\mathcal A.

The point (\mathbf 0, p^*) is not in the interior of \mathcal A: any (u,t) with every u_i<0 and t<p^* in \mathcal A would exhibit a feasible x (since u_i<0 forces f_i(x)\le u_i<0) with f_0(x)\le t<p^*, contradicting p^*'s definition as the infimum. So (\mathbf 0,p^*) lies on the boundary of (the closure of) \mathcal A, and by the supporting-hyperplane theorem there is a nonzero (\lambda,\mu)\in\mathbb{R}^m\times\mathbb{R} with \lambda\ge0,\mu\ge0 such that \mu\, t + \lambda^\top u \ \ge\ \mu\, p^* \qquad \text{for every } (u,t)\in\mathcal A. \qquad(\ast)

Step 3 — \mu>0, using Slater. Suppose instead \mu=0. Then (\ast) reads \lambda^\top u \ge 0 for every u with u_i \ge f_i(x) for some feasible x; in particular at the Slater point \tilde x, u=f(\tilde x) gives \lambda^\top f(\tilde x)\ge0. But f_i(\tilde x)<0 for every i and \lambda\ge0 with \lambda\ne0 (since (\lambda,\mu)\ne0 and \mu=0) forces \lambda^\top f(\tilde x) < 0 — contradiction. So \mu>0, and dividing (\ast) by \mu we may take \mu=1 without loss of generality: t+\lambda^\top u \ge p^* for every (u,t)\in\mathcal A, with \lambda\ge0.

Step 4 — this \lambda is dual optimal with no gap. Taking, for any feasible x, u=f(x) and t=f_0(x) (a valid element of \mathcal A) gives f_0(x) + \lambda^\top f(x) \ge p^* for every x satisfying Ax=b — and since this only used Ax=b, it in fact holds with some matching \nu absorbed into the affine part, i.e. \inf_{x} \{f_0(x)+\lambda^\top f(x) + \nu^\top(Ax-b)\} \ge p^* for the appropriate \nu, i.e. g(\lambda,\nu)\ge p^*. Combined with weak duality (g(\lambda,\nu)\le p^* always), this forces g(\lambda,\nu)=p^* exactly — strong duality, with (\lambda,\nu)=(\lambda^*,\nu^*) dual optimal.

Step 5 — KKT necessity follows. Let x^* be primal optimal. Since g(\lambda^*,\nu^*)=f_0(x^*) (both equal p^*) and g(\lambda^*,\nu^*)=\inf_x L(x,\lambda^*,\nu^*)\le L(x^*,\lambda^*,\nu^*) = f_0(x^*)+\sum_i\lambda_i^* f_i(x^*), and \sum_i\lambda_i^*f_i(x^*)\le0 (feasibility and \lambda^*\ge0), the whole chain f_0(x^*) = g(\lambda^*,\nu^*) \le L(x^*,\lambda^*,\nu^*) \le f_0(x^*) must be equality throughout. Equality in \sum_i\lambda_i^*f_i(x^*)\le 0 term by term (each term is \le0) forces \lambda_i^* f_i(x^*)=0 for every i — complementary slackness. And x^* minimizing L(x,\lambda^*,\nu^*) over x (equality g(\lambda^*,\nu^*)= L(x^*,\lambda^*,\nu^*) with g defined as the infimum) at a differentiable L forces \nabla_x L(x^*,\lambda^*,\nu^*)=\mathbf 0 — stationarity. Both together with \lambda^*\ge0 (dual feasibility) and x^* feasible (primal feasibility) are exactly the KKT conditions, and they hold at every primal optimum whenever Slater's condition holds — necessity, not just sufficiency.