The real quick answer to my question is that the normal ordered form of $T'(w)$ in the statement of my question is exactly right, but the primes above the fields $\phi'(w)$ need to be taken seriously. The meaning is that $\phi'(w)=\phi(z)$, and after making this substitution (noting that we do not change anything about the commutator subtracted in the normal ordering) we indeed see that the expectation value of $T'(w)$ is nonzero and what it should be.
This answer is maybe a little too quick, because when I asked my question I was still a little unsure that $T'(w)$ had the correct form. So let me go through a slightly longer reasoning so that there are no doubts.
Starting on the plane, with coordinate $z$, we have an uncontroversial definition of normal ordering $$:\partial_z\phi(z)\partial_{z'}\phi(z'): \quad=\quad \partial_z\phi(z)\partial_{z'}\phi(z')-\langle\partial_z\phi(z)\partial_{z'}\phi(z')\rangle $$ The first term is a product of two primary fields at different points so we know exactly how it transforms. The second term is not an operator at all (or rather proportional to the identity operator) so we know exactly how it transforms.
There are different ways to think about the transformation, in particular we could think about field configurations in the path integral, but I find it helpful to think in terms of operators. From an operator perspective there will be some unitary operator $U$ associated with the transformation which can ultimately be written in terms of $T(z)$ or the generators of the Virasoro algebra. 364
Its effect on the scalar field is simple. $$U\phi(z)U^{-1}=\phi(w)$$ Note this is the exact same field appearing on both sides, only the argument has changed. So writing our transformation law: $$U:\partial_z\phi(z)\partial_{z'}\phi(z'):U^{-1} =\partial_z w(z)\partial_{z'}w(z')\partial_{w}\phi(w)\partial_{w'}\phi(w')-\langle\partial_z\phi(z)\partial_{z'}\phi(z')\rangle$$ Now since $\phi(w)$ is indeed the same operator, we can rewrite this expression in terms of normal ordered $:\partial_{w}\phi(w)\partial_{w'}\phi(w'):$ the ordinary way. When we take the limit where $z, z'$ approach each other we get the extra central charge term on the left hand side, following the exact same reasoning as <PERSON> et al do in their textbook at (5.136).
Moving the $U$ to the other side we have $$T(z)=\frac{1}{z^2}\left(\left(\frac{L}{2\pi}\right)^2 U^{-1}T(w)U +\frac{c}{24}\right)$$ Now the key thing from the operator point of view is that when we take the vacuum expectation value $\langle U^{-1} T(w)U\rangle$ is not the same as $\langle T(w)\rangle$ which vanishes. They would be the same thing if $U$ were a global conformal transformation, since that would leave the vacuum invariant, but this particular transformation will involve Virasoro generators which do not annihilate the vacuum.
So the non-invariance of the vacuum is really what opens up the possibility of a non-vanishing $\langle U^{-1}T(w)U\rangle\equiv \langle T'(w)\rangle$. And not at all coincidentally the conformal transformations which do change the vacuum are those that have non-vanishing Schwarzian derivative, and thus an extra inhomogeneous central charge term. 818