How can an up-lifting tail have normal force direction and gradient when generating UP force?
Let me break it down in sections. It's rather long I'm afraid.
1. Tail lift direction
Consider the following diagram of a wing-body and tail combination, with a non-pitching aircraft in steady-state. Let's assume that a true wingbody mean aerodynamic centre ($h_{n_{wb}}\overline{c}$) exists and therefore the pitching moment does not vary with AOA at that point.

Image ref: Etkins, Dynamics of Flight
The pitching moment for the whole aircraft about every point must be zero. Let's pick the point of CG ($h\overline{c}$) to sum our moments, since it also corresponds to the point of rotation for a free-body (this equation can be found in Etkins, Dynamics of Flight, but its derivation is simple and you should be able to deduce it from first principle):
$$C_m = C_{m_{ac_{wb}}} + C_L(h-h_{n_{wb}}) - \frac{\overline{l}_t S_t}{\overline{c} S} C_{L_t} = 0$$
where $C_m$ is the total pitching moment, $C_{m_{ac_{wb}}}$ is the wingbody pitching moment coefficient about the MAC, $C_L$ is the total lift coefficient (wingbody + tail), $C_{L_t}$ is the tail lift coefficient normalized over the tail surface area ($S_t$), $\overline{l}_t$ is the distance from the tail MAC to the wing MAC, and $\overline{c}$ is the wing reference chord.
The sign of the tail lift, $C_{L_t}$, to make the above hold can be either positive or negative, depending on the relative contribution of each of preceding terms. Note two facts:

For a positively cambered wing, $C_{m_{ac_{wb}}}$ is typically negative.
$C_L$ is positive for 1G flight.

Consider some extremes:

If the CG is way ahead of the wing MAC ($h \ll h_{n_{WB}}$), then the second term will be very negative, and $C_{L_t}$ must be negative to compensate. That is, negative tail lift.

If the CG is behind the wing MAC ($h &gt; h_{n_{WB}}$), and $C_L$ is very large, then $C_{L_t}$ must be positive to compensate. That is, positive tail lift.
Furthermore, if the CG is behind the wing MAC, then there will always be an AOA above which the trimmed tail lift becomes positive, assuming stall does not occur.


2. Neutral point
Neutral point ($h_n\overline{c}$) is the longitudinal position at which the aircraft does not experience any pitching moment with a variation in AOA or $C_L$ (in another word, NP = MAC of the aircraft). If the CG is ahead of the NP, the aircraft will experience a pitch down moment with increasing lift (statically stable); and vice versa for unstable.
I will cite the following result, once again from Etkins (but simplified):
$$h_n = h_{n_{wb}} + \frac{a_1}{a} \frac{\overline{l}_t S_t}{\overline{c} S} \left ( 1 - \frac{\partial \epsilon}{\partial \alpha} \right )$$
where $a=\frac{\partial C_{L}}{\partial \alpha}$ is the total lift slope of the aircraft, $a_1=\frac{\partial C_{L_t}}{\partial \alpha_t}$ is the isolated lift slope of the tail, $\epsilon$ is downwash from the wing on the tail.
Note that the second term is always positive. Therefore, the CG can be behind the wing MAC, yet ahead of the NP.
3. Elevator hinge moment and stick force
For a reversible aircraft where the column is directly connected with an elevator, stick force gradient is directly conferred by the aerodynamic hinge moment on the elevator (let's assume there is no down-spring or bob-weight for simplicity).
For a non-cambered tail, there are two main contributions:

Hinge moment due to flow incidence ($b_1$)
Hinge moment due to elevator deflection ($b_2$)
Hinge moment due to trim tab ($b_3$)

You are right to point out that if a tail is lifting, and its flow incidence increases with aircraft AOA, then there will be increasingly TE up HM on the elevator, which would imply instability. However, static stability ensures that there is increasing TE up elevator needed to trim for increasing AOA, which will add TE down HM on the elevator.
The details are more nuanced than the above, and it will involve airspeed since HM is squared to airspeed, but it can be shown that the stick force reversal only happens when the CG is aft of the stick-free neutral point ($h_n^{'} \overline{c}$), and not at neutral point:
$$h_n^{'} = h_n - \frac{b_1}{b_2}K$$
where $K$ is a positive constant that is a function of the aircraft geometry (I've simplified from eqn 2.6.13 from Etkins). As you can see, if $b_2$ (HM due to elevator deflection) overpowers $b_1$ (HM due to flow incidence on tail), then we pretty much recover the NP. Down-spring and anti-servo tabs are some methods to increase the $b_2$ artificially.
More specifically, please consult the following figure, adapted from Bromfield, Factors affecting the apparent longitudinal stick-free static stability of a typical high-wing light aeroplane and Etkins, Dynamics of Flight:

In the graph, $\delta_t$ is the trim tab deflection.
Observations:

The first two curves are with the tab always set to 0 incidence against the elevator. The movement of the CG purely shifts the stick force vs. V curve up and down. When the CG is aft of the stick-free neutral point ($h&gt;h_n^{'}$), any steady-state speed would require a pull force (despite the stick may be forward of neutral if the CG is still ahead of the neutral point!)

The last two curves are with the tab set for a trim speed ($V_{trim}$). Now, when the CG is aft of the stick-free neutral point, the stick force gradient also goes unstable! What effectively happens when the CG coincides with the stick-free neutral point is that the hinge moment contribution from the elevator+tab cancels that from the tail AOA; therefore, zero steady-state force is needed to hold any particular elevator/stick position.


Either way, whether the tail is lifting doesn't dictate the stability of the stick force vs. airspeed.