Dimension of $M_2(\Gamma_0(14))$ The dimension of $M_2(\Gamma_0(14))$, according to LMFDB, is 4. This should be decomposed into oldforms and newforms, and indeed - the dimensions are 1 and 3, respectively.
According to my understanding, the oldforms should come from spaces with level dividing 14, that is 2 and 7 (and there are none of level 1). However, $\dim(M_2(\Gamma_0(7))=1$ and $\dim(M_2(\Gamma_0(2))=1$, so one oldform is "missing". Where have I made a mistake?
 A: Note: In my answer I will simplify notation, writing $M_2(N)$ to mean $M_2(\Gamma_0(N))$.
You are correct that $\dim(M_2(7))=\dim(M_2(2))=1$. 
These spaces contain the Eisenstein series of weight 2 as described in section 3.5 of these notes (http://www.few.vu.nl/~sdn249/modularforms16/Notes.pdf). In particular, 
$$E_2^{(7)}(z)=E_2(z)-7E_2(7z)\tag{1}$$
is a modular form in $M_2(7)$ and
$$E_2^{(2)}(z)=E_2(z)-2E_2(2z)\tag{2}$$
is a modular form in $M_2(2)$. These are oldforms as you've stated and together span a 2-dimensional subspace of $M_2(14)^{\text{old}}$. However, the definition of oldforms also encompasses modular forms of the form $f(dz)$ where $f(z)$ is a modular form at a level $M$ strictly dividing $N$ and $d$ divides $N/M$. Hence, the forms $E_2^{(7)}(2z)$ and $E_2^{(2)}(7z)$ are also oldforms. Explicitly we have
$$E_2^{(7)}(2z)=E_2(2z)-7E_2(14z)\tag{3}$$
and
$$E_2^{(2)}(7z)=E_2(7z)-2E_2(14z)\tag{4}.$$
The forms (1), (2), (3) and (4) form a spanning set for $M_2(14)^{\text{old}}$. Although they are linearly dependent, any 3 of them can be used to form a basis for $M_2(14)^{\text{old}}$. This explains why $\dim(M_2(14)^{\text{old}})=3$.
