Question: [![enter image description here](https://i.stack.imgur.com/1bCjE.png)](https://i.stack.imgur.com/1bCjE.png)

In the blue circle, on the left of the equation. What **base** does $\bf u'$ have here? If I actually calculate $\bf u'$ what will the vector I get stand for?

Am I going to get the vector $\bf u'$ relative to the standard unit vectors $(1,0), (0,1)$ and it's just invisible since it's kinda the default bases?

Answer: 
The answer as to values of the basis vectors $u\_i'$ that compose the basis $B'$, is that they completely depend on the vectors which define your "original" basis $B$. **The values of the $u\_i'$ thus calculated correspond to the coordinates of the $u\_i'$ vectors, expressed in basis $B$**.

That's for the "raw" answer to your question. But let me discuss a bit for your benefit.

Generally, indeed, $B$ is the "canonical" basis (with each vector having a $1$ in some coordinate, and $0$s everywhere else). But this in no way is forced to be the case.

However, the formula for change of basis IS universal: that means that this formula works no matter the $B$ and $B'$ involved, so long as you know the way the basis vectors of $B'$ are expressed in basis $B$'s coordinates (that is, your $[a \space b]^T$ and $[c \space d]^T$ above, which define the change-of-basis matrix), you can calculate your new basis vectors.

There is one tricky thing though: expressing the basis vectors of $B$, in the $B$ coordinate system, will ALWAYS make it seem like $B$ is the canonical basis ! Why ? Because the first basis vector of $B$ will have coordinates $(1, 0, 0, 0, ..., 0)$ in basis $B$, etc. No matter the basis $B$. Another way to phrase this same idea: the change-of-basis matrix from $B$ to $B$ is obviously the identity matrix, as nothing changes. And the column vectors of the identity matrix is precisely what the canonical basis "looks like".

The distinction you have to make is "what is the geometry of my basis vectors ?". If they're not all length 1, nor all 2-by-2 perpendicular with each other, your basis $B$ is quite probably not the canonical basis (which has to be orthonormal).

The reason for the confusion, is that when you work in a purely algebraic context (like most textbooks or exercises do), you have to "build up" your geometry from algebraic symbols, and expressing your basis $B$ in the canonical basis (so that you can know the geometry of the vectors of $B$) is sort of the only way to do things intuitively (ie, transform algebra into geometry through the use of number coordinates, by relying on the convention that you'll always start from the orthonormal 'canonical basis').

Do check out 3blue1brown's YouTube series "Essence of Linear Algebra", where he explains (among other things) change of basis in an intuitive, visual way.

PS: Do note that when I say "the formula for change of basis is universal", there is a slight caveat. Notions of covariance vs contravariance enter into play later on. I suggest the YouTube channel eigenchris to see what this means when you get there.

In a nutshell, if you want your (column-type) vector to be an absolute, an "invariant", then increasing the length of your basis vectors will force you to reduce the value of the respective coordinates, so that the shape of your vector stays the same "geometrically" speaking. Changing "the same way as the basis vectors" is called covariance, changing "the opposite way as the basis vectors" is called contravariance (and you'll use the matrix inverse of your change-of-basis matrix).

This is the other way around for (row-type) covectors. Basis covectors are contravariant, and covector coordinates are covariant.
Let $V$ be a vector space. Let it be 2-dimensional euclidean space, i.e. the plane of a black/white board for concreteness. Draw 2 non-parallel lines on it. Call one the x-axis, the other y-axis. You are free to choose any orientation for these 2 lines except when they are parallel.

Draw an arrow in any direction of any length. Drop lines parallel to axes from the tip of the arrow onto the axes and measure off the distances from origin and note the directions of the points where the lines dropped meet the axes. The distances are respectively the components of the vector represented by the arrow. This vector exists independently of it's representation in terms of components. It's components depend on the direction of the two lines you chose initially as axes of the plane.

Choose two perpendicular lines as axes. The unit vectors along the axes will be $(1,0)$ say $u\_1$, the unit vector in the x-direction and $(0,1)$ say $u\_2$, the unit vector in the y-direction. These two serve as the basis for the plane.

The set $\{(0,1),(1,0)\}$ is linearly independent. Any vector $(x,y)$ in this space can be represented as $x\cdot(1,0)+y\cdot(0,1)$.

Another choice for basis can be $\{(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}), (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})\}$ which we denote as $u\_1^{'}$ and $u\_2^{'}$. The new basis vectors are also elements of the vector space and can be expressed as linear combinations of $u\_1$ and $u\_2$:

$$\{\frac{1}{\sqrt{2}}\cdot u\_1+\frac{1}{\sqrt{2}}\cdot u\_2,\frac{-1}{\sqrt{2}}\cdot u\_1+\frac{1}{\sqrt{2}}\cdot u\_2\}$$

Even the old basis can be expressed in terms of the new one as:

$$\{\frac{1}{\sqrt{2}}\cdot u\_1^{'}+\frac{-1}{\sqrt{2}}\cdot u\_2^{'},\frac{1}{\sqrt{2}}\cdot u\_1^{'}+\frac{1}{\sqrt{2}}\cdot u\_2^{'}\}$$

Whenever working with representations of vectors, stick with one base. Substitute values of components in above expressions to check that above formulae are correct.