$$\ ϕ_{t*}\biggl(\frac∂ {∂x^j}\bigg|_ p \biggr) = \sum_i \frac{∂ϕ^i} {∂x^j} (t, p) \frac∂ {∂x^i}\bigg|_{ ϕ_t(p)}$$ Thus, if $\ Y = ∑b^j\frac∂{∂x^j}$ , then $$\ ϕ_{−t∗} Y_{ϕ_t(p)} = \sum_j b_j (ϕ(t, p))ϕ_{−t∗} \biggl(\frac{∂}{∂x^j}\bigg|_{ϕ_t(p)}\biggr)= \sum_{i, j} b_j (ϕ(t, p))\frac{∂ϕ^i}{∂x^j} (−t, "p") \frac∂{∂x^i}\bigg|_p $$
Shouldn't the "p" part be $\ \phi_t(p) $ instead of p?
The answer is yes. Good catch!