What's the physical meaning of the standard base vectors?

Your basis vectors are unit vectors that are (ironically) unitless. So for all of the examples of basis vectors you propose, specifying what the vector represents is somewhat misleading.

In other words, you can think of all of your basis vectors ($\hat d_1$, $\hat d_2$, $\hat d_3$, $\hat F_1$, $\hat F_2$, $\hat F_3$, $\hat i$, $\hat j$, $\hat k$) as just directions in space. These sets of basis vectors do not represent positions, forces, etc. Your actual component magnitudes ($a$, $b$, $c$, etc.) have units, and this is what determines what the vector physically represents.

This is why you can explain many vector quantities with the same set of basis vectors. Position, force, etc. all have some direction in the space you are looking at. Therefore, you can express each vector using the same basis vectors.

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So, when you say, "I can find three displacement vectors and use them as basis vectors," what you really mean is, "I have three displacement vectors, and I can take their directions and define basis vectors." (assuming they actually form a valid basis). With your example, these are $$\hat d_1=\frac{\mathbf d_1}{|\mathbf d_1|}$$ $$\hat d_2=\frac{\mathbf d_2}{|\mathbf d_2|}$$ $$\hat d_3=\frac{\mathbf d_3}{|\mathbf d_3|}$$

These basis vectors are just directions, they have no units associated with them, and you can use them to explain any other vector. For example, a force could be $$\mathbf F=F_1\,\hat d_1 +F_2\,\hat d_2+F_3\,\hat d_3$$ In other words, you have broken your force vector into components where each component is along the direction of each of displacement vectors you used to define your basis. The values $F_i$ have units of force, and this is what makes the vector a force vector. Also, note, for example, that the term $F_1\,\hat d_1$ is not "changing the intensity" of $\hat d_1$. It is just a scalar multiplication of the unit vector $\hat d_1$. This is analogous to how computing $3\cdot 4=12$ does not change what $4$ actually is.