Suppose the linear transformation $f:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ with the following conditions:
$\dim(f(\mathbb{R}^n))=k$
$k<\min{(m,n)}$
Show that there is a base like $\mathfrak{B}$ for $\mathbb{R}^n$ and a base like $\mathfrak{B}^\prime$ for $\mathbb{R}^m$ such that:
$M^{\mathfrak{B}}_{\mathfrak{B}^\prime}(f)=\begin{bmatrix} I_k & O\\O&O \end{bmatrix}$
I tried to find $\ker(f)$ dimension $n-k$, But I can't find the proper base. I think i should use $\ker(f)$ elements but I have no idea how to do that?
Since there is no specific transformation given, you won't be able to actually give specific elements of the bases, but you can describe how you would be able to do this - in general.
As you said, the dimension of the kernel is $n-k$. Since the kernel is a subspace of $\mathbb{R^n}$, you can find a basis of this kernel and it will consist of $n-k$ elements: $\left\{b_{k+1},\ldots,b_{n-1},b_n\right\}$.
You can then expand this to a complete basis of $\mathbb{R^n}$ by adding $k$ vectors, making sure all vectors are linearly independent, ending up with a basis $\mathfrak{B}$ of the form $\left\{\color{green}{b_1,\ldots,b_k},b_{k+1},\ldots,b_n\right\}$.
Recall that the $i$th column in the matrix representation $M^{\mathfrak{B}}_{\mathfrak{B}^\prime}(f)$ are the coordinates of the $i$th basis vector of $\mathfrak{B}$ (with respect to the basis $\mathfrak{B'}$).
Can you see how this choice of $\mathfrak{B}$, regardless of which basis $\mathfrak{B'}$ for $\mathbb{R^m}$ you pick, already realizes 'half' of the demanded form of the matrix, namely the parts in blue:
$$M^{\mathfrak{B}}_{\mathfrak{B}^\prime}(f)=\begin{bmatrix} \color{red}{I_k} & \color{blue}{O}\\ \color{red}{O} & \color{blue}{O} \end{bmatrix}$$
Maybe you can take it from there?
Hint: use the images of the basis vectors of $\mathfrak{B}$ to construct a basis $\mathfrak{B'}$ that realizes the parts in red above.