Answer My Own Question:
The short answer:
We don't have to verify $$\forall 1 \le i \le m: \Phi(T_i) \ge \Phi(T_0).$$ This condition is a special case of a more general one, which we need to verify for this example.
The detailed answer:
Consider a data structure $D$ and its states $D_0 \cdots D_m$ produced by a sequence of $m$ operations $o_1 \cdots o_m$:
$$
D_0,\; o_1,\; D_1,\; o_2,\; \cdots,\; \underbrace{D_{i-1},\; o_{i},\; D_{i}}_{\text{the $i$-th operation}},\; \cdots,\; D_{m-1},\; o_{m},\; D_{m}.
$$
The amortized cost of the $i$-th operation $o_i$ is defined as:
$$\hat{c_i} = c_i + \Big(\Phi(D_{i}) - \Phi(D_{i-1})\Big)$$
Therefore, the total actual cost of these $m$ operations equals the total amortized cost plus the net decrease in potential from the initial to the final state.
$$
\sum_{1 \le i \le m} c_i = \left( \sum_{1 \le i \le m} \hat{c_i} \right)
+ \Big(\underbrace{\Phi(D_{0}) - \Phi(D_m)}_{\text{net decrease in potential}} \Big).
$$
For each upper bound $\Box$ for $\Phi(D_{0}) - \Phi(D_m)$, we get a corresponding upper bound for the total actual cost of all operations:
$$
\sum_{1 \le i \le m} c_i \le \left( \sum_{1 \le i \le m} \hat{c_i} \right) + \Box.
$$
The condition $$\forall 1 \le i \le m: \Phi(D_i) \ge \Phi(D_0)$$ in the post is just a special case of deriving upper bounds for the total actual cost, where the upper bound is the total amortized cost. Moreover, it is common to choose $\Phi$ such that $\Phi(D_0) = 0$ and $\Phi(D_i) \ge 0$ for all $i$.
To verify the validity of the potential function for splay trees, we use the following upper bound for $\Phi(T_{0}) - \Phi(T_i)$:
$$\Phi(T_{0}) - \Phi(T_i) \le n \log n.$$
