I hope that this answer gives you some intuition. Firstly, consider what it means to be diagonal in the first place. Say we have a diagonal transformation:

$$ \begin{pmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 & \lambda_2 & \dots & 0 \\
\vdots & \ddots & \dots \\
0 & \dots & \dots & \lambda_n
\end{pmatrix}$$
Now what does this actually mean? Well, I am sure that you can see that it means when viewed in terms of **some** basis of $\Bbb{R}^n$ (or whatever v. space we're talking about, I went for $\Bbb{R{^n}}$ as it is most intuitive) we simply rescale the axes of the basis. Mapping $(1, 0, \dots, 0) \mapsto (\lambda_1, 0, \dots, 0)$

$(0, 1, 0, \dots, 0) \mapsto (0, \lambda_2, 0, \dots, 0)$ etc.

In this regard, actually, we see that being defective should be considered the more '*normal*' case. Being a linear transformation is simply a much weaker condition than having to be a transformation which re-scales the axes (under some identification of axes in $\Bbb{R}^n$).

For example consider the *skew*-transformation (purposefully written in italics):

$$\begin{pmatrix}
1 & 1 \\
0 & 1 \end{pmatrix}$$

This is the transformation skewing the plane, sending $(0, 1) \mapsto (1, 1)$

We send the square with vertices $(0,0), (0,1), (1,1), (1,0)$ to the parallelogram with vertices $(0,0), (1,1), (2,1), (1,0)$:

Now, I don't know if you know the criterion for being diagonalizable, it's whether or not the minimal polynomial (look this up if you like) splits in to distinct linear factors over the field that you're working in.
But this is kind of unimportant for our intuition. The characteristic polynomial of this matrix is $(1-x)^2$ and the minimal polynomial happens to also be $(1-x)^2$ so we don't have **distinct** linear factors and this matrix is not diagonalizable.

But **why** is this the case?? Well, let's pick axes in $\Bbb{R}^2$. We can pick the horizontal $x$-axis. -This maps to itself so is rescaled by $1$ (or not rescaled at all). Looking good for diagonalization so far!!. But now we have to pick another axis to cover $\Bbb{R}^2$. Ahhhhhhhhhh...

Our second axis that we pick has to have some vertical component (we can't pick the $x$-axis again, as this wouldn't cover $\Bbb{R}^2$ fully, we'd just have the $x$-axis!!). However picking any axis here with some vertical component causes a problem. Say we pick the line $y=x$ as our second axis, we see that this line gets skewed horizontally (to the line $y=\frac12x$). No matter what second axis we pick the vertical component gets skewed horizontally and this transformation doesn't simply re-scale the line. So we can't be a re-scaling transformation, no matter what axes of $\Bbb{R}^2$ we pick!!! So we can't be diagonalizable.

In the above depending on how much linear algebra you know replace words "axes" with "basis" and not picking the $x$-axis twice is due to the fact that we need linear independence.

What I hope that this makes obvious is that fact that actually, being diagonal is the very special property. Not being defective, which is the more normal case! Hopefully this fills in your intuition a bit. Don't hesitate to ask any further questions!