Given that you already accept that the duality gap along the central path is $m/t$, then the inequality you're struggling with is really rather simple. Remember, the dual problem provides lower bounds for the optimal value of the primal. So $p^*$ is necessarily in between the objective values of any feasible primal point and any feasible dual point.
That is: we have
$$f(x^*(t)) - g(\gamma^*(t),\lambda^*(t)) = m/t$$
where $g$ is the dual objective. But we also have
$$p^* \leq f(x) \quad \forall x\text{ feasible}$$
and
$$p^* \geq g(\gamma,\lambda) \quad \forall \gamma,\lambda\text{ feasible}$$
Thus for any feasible primal and dual points, we have
$$f(x)-p^* \leq f(x) - g(\gamma,\lambda) \quad \forall x,\gamma,\lambda\text{ feasible}$$
And along the central path,
$$f(x^*(t))-p^* \leq f(x^*(t)) - g(\gamma^*(t),\lambda^*(t))=m/t.$$
EDIT: Actually, it looks like you may not accept the premise that the $m/t$ difference applies to the original problem. So let's look at this for an LP. The primal and dual problems are
$$\begin{array}{llcll}
\text{minimize} & c^T x & \quad & \text{maximize} & b^T \lambda \\
\text{subject to} & A x = b & & \text{subject to} & A^T \lambda + \gamma = c \\
& x \succeq 0 & & & \gamma \succeq 0
\end{array}$$
I prefer $y$ and $z$ above to $\lambda$ and $\gamma$, but for consistency I am keeping your variable names. $\lambda$ is the Lagrange multiplier for the equality constraints, $\gamma$ is the Lagrange multiplier for the inequalities. The barrier problem for the primal is
$$\begin{array}{ll}
\text{minimize} & t c^T x - \textstyle \sum_{i=1}^m \log x_i \\
\text{subject to} & A x = b
\end{array}$$
Note that the domain of this problem is limited to $x\succ 0$ by the barrier term.
The optimality conditions for a fixed $t>0$ satisfy
$$t c - \mathop{\textrm{diag}}(x)^{-1} \mathbf{1} - A^Ty = 0 \quad Ax=b \quad x \succ 0$$
where $y$ is a Lagrange multiplier for the equality constraints. That second term is a bit clumsy: define $z_i\triangleq x_i^{-1}$ for $i=1,2,\dots,m$ so
$$t c - z - A^Ty =0 \quad Ax=b \quad x,z \succ 0 \quad z_i\triangleq x_i^{-1},~i=1,2,\dots,m$$
By inspection, we see that if $(x,y,z)$ satisfies these equality constraints, then $(\gamma,\lambda)=(t^{-1}z,t^{-1}y)$ is a feasible dual point for the original problem. The duality gap for this primal/dual pair $(x,\gamma,\lambda)$ is
$$c^Tx-b^T\lambda=c^Tx-t^{-1}b^Ty=c^Tx-t^{-1}x^TA^Ty=(c-t^{-1}A^Ty)^Tx=t^{-1}z^Tx=m/t.$$
So even though we're looking at the modified barrier model, the set of optimal points $x^*(t)$ are all feasible for the original, and they all lead to feasible dual points $(\gamma^*(t),\lambda^*(t))$ for the dual problem as well. The duality gap for the original model is $m/t$.
