Let $F:=\mathbb{F}_{q^m}$ be an extension field of the finite field $\mathbb{F}_q$. A polynomial $L(x)$ is called a linearized polynomial over $F$ if $L(x)$ is of the form $$ L(x)=\sum_{i=0}^d a_i x^{q^i} $$ with $a_i\in F$ for all $i$. Now suppose $K$ is an extension field of $F$. The map taking $\beta \in K$ to $L(\beta)$ is linear, so for $\beta_1,\beta_2\in K$ and $\lambda_1, \lambda_2\in \mathbb{F}_q$ it holds that $L(\lambda_1 \beta_1+\lambda_2 \beta 2)=\lambda_1 L(\beta_1)+\lambda_2 L(\beta_2)$.
My first question is how do interpret $L(\beta)$? Isn't $\beta$ vector valued so how do even calculate $\beta ^{q^i}$?
Furthermore I know that the kernel of $L(x)$ is a subspace of $K$, so a vector space over $\mathbb{F}_q$. Now my textbook states that
If $\deg(L(x))=q^d$, then this vector space has dimension at most $d$ but the dimension could be smaller if $L$ has repeated roots.
I'm not really sure why this is true?
1) You can see $\beta$ as a vector over $\mathbb F_q$ if you fix a basis $\alpha_1, \ldots, \alpha_m$, where $\alpha_i \in F$. Hence you can write $$ \beta=\sum_{i=1}^m c_i\alpha_i , \;\;\;\; c_i \in \mathbb F_q.$$ But $\beta$ is an element of $K$, so the evaluation of $L$ in $\beta$ is exactly what you expect, i.e. $$L\left( \sum_{i=1}^m c_i\alpha_i\right).$$
2) This is true because $L(x)$ is a polynomial of degree $q^d$, hence the number of its roots (counted with multiplicity) is $q^d$. Since the kernel of $L$ is exactly the set of roots of $L(x)$, it has at most $q^d$ elements. Moreover it is a subspace, so its dimension as a vector space is at most $d$. In general the dimension of the kernel of $L$ is given by $d-a$, where $$ a:=\min\{i\;|a_i \neq 0 \}.$$