I have the following definition:
Laplace transform of a random vector $X$ is the map $z \rightarrow E\exp(z^T X)$
I don't understand what is $z$ here. I guess it is some vector. However why does the LT map $z$ into something, rather than $X$ into something? Why isn't the definition like this : $X \rightarrow E\exp(z^T X)$ for some vector $z$? I guess I'm missing something here, maybe a concrete example would help
As you have written, $X$ is a random variable. This means that $X$ takes values distributed according to a certain probability law. A random variable cannot be used as an independent variable as you have proposed, because it takes unpredictable values.
Let us take an example with a vector of dimension 1. So $X$ is a one-dimensional random vector, it is a random number and let's say it can take the value $1$ with probability $1/2$ and the value $0$ with probability $1/2$. Let us compute the map $z\mapsto \mathbb{E}[\exp(zX)]$. For any complex number $z$, we have $$ \mathbb{E}[\exp(zX)]=\frac12\exp\left(z\times1\right)+\frac12\exp\left(z\times0\right)=\frac{\mathrm e^{z}+1}2.$$ The Laplace transform of $X$ is the function $\phi:z\mapsto\frac{\mathrm e^{z}+1}2$.
Generally speaking, the Laplace transform of a random variable is a function that contains all possible informations concerning $X$. One always has $\phi(0)=1$. One can extract from it the mean value of $X$ as $$\overline{X}=\phi'(0)=\frac12$$ and many other properties. It is handier than manipulating probabilities and also more efficient.