What the primed quantities really are in this context? In this definition:

The set of $N$ quantities $V_j$ is said to be the components of an $N$-dimensional vector $\mathbf{V}$ if and only if their values relative to the rotated coordinate axes are given by
$$V_i'= \sum_{i=1}^N a_{ij}V_j, \quad i=1,2,\dots,N$$
As before, $a_{ij}$ is the cosine of the angle between $x_i'$ and $x_j$. Often the upper limit $N$ and the corresponding range of $i$ will not be indicated. It is taken for granted that you know how many dimensions your space has.

a vector is characterized as a set of components $\{V_j\}$ such that relative to some other coordinate system these components are given by that formula.
This extends to covectors and tensors of all ranks.
The problem is: I believe this definition assumes we have $V_j$, we have $V_i'$ and we can judge whether or not the relation holds. If it holds, $V_j$ are the components of a vector.
But how on earth are $V_i'$ being defined if not by that equation?
What I learned in linear algebra was: given a vector space $V$ and a vector $v\in V$ if we have a basis $\{e_i\}$ we can write uniquely $v = \sum V_i e_i$. This gives the $V_i$.
If we have another basis $\{f_i\}$ we can also write uniquely $v = \sum V_i'f_i$ and this is equivalent to saying
$$v = \sum V_i e_i = \sum V_i a_{ij} f_j$$
and again by uniquenes we have $V_j' = \sum V_i a_{ij}$.
But wait a minute, this means that $V_j'$ is equal to that by definition. In other words, given $V_i$, there is no other option for $V_j'$. This would mean that given any set of components there could be no way to fail this condition.
But I'm obviously missing the point. If the quantities $V_i'$ were defined like this for all sets of components $V_i$ there would be no reason to check if the transformation law is obeyed.
In that context: given a set of numbers $V_i$ how the primed quantities $V_i'$ are defined in order to check that $V_i$ is a vector? Given components $V_i$, what physicists mean by $V_i'$?
 A: Following up my comment, I would like to present an example. Consider a basis $(e_1, \dots, e_n )$ and also a basis $(f_1,\dots , f_n) $. Then there is a unique set of numbers such that $f_i=T^j_i e_j$. If using summation convention some vector $V=v^i e_i=w^jf_j$ then as you pointed out $$v^i e_i=w^j T^i_j e_i $$ and hence $v^i=T^i_j w^j $ by linear independence. This is the ttansformation rule for vector components. Now let's apply the definition above. Suppose $\{a^i \}$ and $\{b^i\} $ are components of vectors. Then by your definition $X^i:=a^i+3b^i $ are also the components of a vector (I know it's stupid but as long as you don't have derivatives of vector-FIELDS this is about the most general thing you can do).
The proof:
Transform constituents: $$a'^i+3b'^i=T^i_j a^j+3T^i_j b^j=T^i_j (a^j+3 b^j)=T^i_j X^j=X'^j $$
Hence transforming the individual constituents of the expression and transforming the whole thing as a vector component give the same answer.
Not-Example: 
$Y^i:=a^i\cdot a^i $ are not the components of a vector. (in your terminology; surely there is still the possibility to malliciously misunderstand this and counter "Any numbers define a vector, what are you talking about?") 
Proof:
Transform constituents $$a'^i a'^i=\sum_{j,k} T^i_j a^j T^i_k a^k\neq T^i_j Y^j. $$
As I said, this only gets interesting if you start talking about vector fields and their derivatives. Exercise: Show that the cross product of two vectors is not a vector, if one is allowed to consider coordinate transformations including reflection (I don't see why your definition would restrict to rotations in the first place)
