Consider the function $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$. Let $A = \{ (1,2,3)^t, (1,0,4)^t,(0,0,2)^t \}$ a base of $\mathbb{R}^3$ and $B = \{ (1,1)^t , (2,1)^t) \}$ a base of $\mathbb{R}^2$. $$ f((x,y,z)^t) := (x,y)^t$$ I calculated the transformation matrix $$ M_B^A(f) = \begin{pmatrix} 3 & -1 & 0 \\ -1 & 1 & 0 \\ \end{pmatrix}$$
Should not the equation $M_B^A(f) \cdot v = f(v)$ hold true? (it does not for $v=(1,2,3)^t$) What am I understanding wrong?
$$M_B^A(f) \cdot v = f(v)$$ does not hold true.
But,
$$ M_B^A(f) \cdot [v]_{A} = [f(v)]_{B} $$
definitely does. Here $[u]_{\scr B}$ denotes the coordinate matrix or the coordinate vector of $v$ with respect to the basis $\scr B$. This is the definition of the transformation matrix. What you say will kind of work if you use the standard bases $ \{(1,0,0), (0,1,0), (0,0,1)\} $ and $\{(1,0), (0,1)\}$. But again you will not get a vector which would be a duple of the form $(x, y)$ but a matrix of the form $ \begin{pmatrix} x \\ y \\ \end{pmatrix} $.