I agree with Mathemagician that the use of the term "reaction" here is not the same as when describing action/reaction forces.
As Mathemagician correctly points out, the normal "reaction" from the ground on the box in your 1st diagram is not the action/reaction partner the weight of the box, because (i) weight is gravitational whereas normal force is electromagnetic, (ii) these 2 forces act on the same body (the box), and (iii) weight (or its normal component) and normal reaction force are not always equal.
The action/reaction partner to the weight of the box (ie the downward pull of the Earth on the box) is the gravitational pull of the box acting upwards on the Earth. And the action/reaction partner to the electromagnetic contact force of the ground pushing up on the box is the electromagnetic contact force of the box pushing down on the ground.
In this context the contact force is called a reaction force because it is the "reaction" or "response" to a given situation and it is difficult to calculate or measure directly. The only practical way that you can decide what value it has is by recognising that it is the unknown response to some other force or forces which are known, in order to make a known acceleration happen. It is the unknown force which is needed to make the equation $$\sum F=ma$$ balance.
For example, if the box weighs $W=20N$ (downwards) and there is no acceleration ($a=0$) then we can deduce that the normal force (upwards) is $N=20N$, to make the equation balance. If the box is on the floor of a lift which is accelerating upwards at $1g$ we can deduce that the contact force must be $N=40N$.
The same can be said of the tangential reaction force, better known as the static friction force. It is another contact force whose value is variable but very difficult to calculate or measure except by deduction from applying the equation $\Sigma F=ma$.
In theory both normal and tangential reaction forces could be calculated directly - eg by modelling them as spring-like forces. Just as you can calculate weight using $W=mg$ if you know the mass $m$ and the strength of the gravitational field $g$, you can also calculate the normal force using $F=kx$ if you know the amount $x$ by which the ground is compressed by the box and the spring constant $k$ for the ground, assuming it obeys Hooke's Law. But it is not practical to measure either $x$ or $k$ because one is microscopically small and the other is enormously large. (Alternatively we could measure the compression of the box, which if made of wood or cardboard will be greater than that of a concrete, granite or marble floor.) Likewise the static friction force could be calculated from the microscopic lateral displacement of the box or the ground, and its enormous lateral spring constant - but again it is not practical to measure either of these.
Mathematically a contact force (normal or tangential) is one type of constraint force. It takes whatever value is necessary to ensure that the constraint condition is fulfilled. Usually this means that a boundary is not crossed.
We assume that the ground and the box are rigid bodies which do not deform at all, not even on an atomic level. This means that $k=\infty$ and therefore $x=0$ for all values of applied force $F$. There is no way of deciding what value $F=kx$ has using this model - it is indeterminate. The only way we can calculate the contact force $F$ is by deduction, as the unknown force in the equation $\Sigma F=ma$. In a 2D problem, if we have 2 unknown forces we can use the rotational equation $\Sigma T=I\alpha$ to find the 2nd unknown force. If we have 3 unknown forces then we're in trouble. The rigid body model cannot cope with this situation.