I'm currently reading a section of a book on an implicit Ruge-Kutta method, and the following is written:
We start, say, with a diagonally implicit Runge-Kutta method $$k_i = hf\left(y_0+\sum\limits_{j=1}^{i-1}a_{ij}k_j + a_{ii} k_i\right), i=1,\dots, s$$ $$y_1 = y_0+\sum\limits_{i=1}^s b_i k_i $$ applied to the autonomous differential equation $$y'=f(y)$$
The main idea is to linearize [the formula above]. This yields $$k_i = hf(g_i) + hf'(g_i) + hf'(g_i)a_{ii}k_i$$ $$g_i = y_0 + \sum_{j=1}^{i-1} a_{ij} k_j$$
Maybe my question is naive but how does one know that the first equation above is not linear while the other one is?
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + \sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes $$x = hf(b + mx)$$ and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols, $$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$ You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.