A Lie algebra associated to a foliation(A kind of saturation of foliations) Inspired  by the  Lie algebra discussed   in this   answer,  we  consider the  following  Lie  agebra  associated to  a  given foliation:
Let  $\mathcal{F}$  be  a nontrivial foliation of  a  manifold  $M$  tangent to  integrable subbundle $D$  of  $TM$.  We define  the  following  Lie  algebra  of  vector  fields on $M$:
$$A_{\mathcal{F}}=\{Y\in \chi^{\infty}(M)\mid [X,Y]\;\; \text{is  tangent to $\mathcal{F}$} \text{ for  all } X\in \Gamma (D)\} $$
In  fact $A_{\mathcal{F}}$ is  the  idealizer of  the  Lie  algebra $L_{\mathcal{F}}$  of  vector  fields on $M$ which are tangent to the  foliation.

First  Question: Is there a  foliation $\mathcal{F}$ for  which  $A_{\mathcal{F}}=L_{\mathcal{F}}$?  
The  second  question:Is it true  to say that the  dimension of  $\{Y(p)\mid Y \in A_{\mathcal{F}}\}\subseteq T_pM$ is  independent of  Choosing $p\in M$? 

If the  answer  of the  second  question is yes, then $A_{\mathcal{F}}$  defines an integrable   distribution $D'$ containing the initial  distribution $D$. It  generates  a  foliation $\mathcal{F}'$ which would be  defined  as  saturation of  $\mathcal{F}$. 

Is there an  example of  this  situation such that $dim(\mathcal{F}) <dim({\mathcal{F}')<dim(M)}$?

Remark: As  we see in the  linked question, when we  have  a  $1$-dimensional foliation $\mathcal{F}$ tangent to  a  non vanishing  vector  field $X$ on  a  surface  $M$ with  volume  form $\omega$  , then the  Lie  algebra $A_{\mathcal{F}}$ is  equal to $$\{Y\mid [X,Y]=fX,\;\text{for  some }f\in C^{\infty}(M)\}=\{Y\in X^{\infty}(M)\mid X.\omega(X,Y)=Div X\omega(X,Y)\}$$.
Added: According to the comment of Bertram Arnold we add the following question:

Is it true to say that there is an open dense subset $U\subset M$ with the following two properties:

1) For every $x\in U,\; \{V(x)|V\in A_{\mathcal{F}}\}=T_x M \}$.
2)$U$ and $M\setminus U$ are $\mathcal{F}$- saturated.
Then it seems that $M\setminus U$ is a characyteristic set in the sence that it is invariant under every leaf preserving diffeomorphism.
 A: 1. First question
The answer to your first question is yes: consider $D=TM$. 
In other words, the foliation $\mathcal F$ has a unique leaf, which is $M$ itself. Hence $A_{\mathcal F}=L_{\mathcal F}=\mathfrak{X}(M)$. 
2. $A_{\mathcal F}$ are not sections of a bundle on $M$
As Bertram Arnold says, $A_{\mathcal F}$ is not stable under multiplication by functions in $C^\infty(M)$. But it is stable under multiplication by functions $f$ that are constant along the leaves (functions on $M/\mathcal F$, so to say). 
3. A bit of symplectic geometry
There is a nice symplectic interpretation of the geometric setup you describe. You can skip it and go directly to the next paragraph if you don't know anything about symplectic geometry. Consider the cotangent space $T^*M$, which is a symplectic manifold. Functions on this are generated by functions on $M$ and vector fields, and the corresponding Poisson bracket is defined by: 
$$
\{X,f\}=X\cdot f\quad\mathrm{and} \quad\{f,g\}=0\,.
$$
On can then consider the ideal $\mathcal I_D$ generated by $\Gamma(D)$. Geometrically this is the ideal of functions vanishing on the zero locus of the induced dual map 
$$
T^*M\longrightarrow D^*
$$
This zero locus is a coisotropic submanifold, and thus one can perform symplectic reduction. Algebraically (i.e. on the level of functions), it amounts to consider the quotient $NP(\mathcal I_D)/\mathcal I_D$ of the Poisson normalizer $NP(\mathcal I_D)$ of the ideal by the ideal itself. 
Reduced spaces are in general quite singular. But, morally speaking, the reduced space shall be thought of as $T^*(M/\mathcal F)$, the cotangent to the leaf space $M/\mathcal F$. 
4. What is $A_{\mathcal F}/L_{\mathcal F}$ ?
One can see from 2 and 3 above that $A_{\mathcal F}/L_{\mathcal F}$ is the Lie algebra of vector fields on $M/\mathcal F$. If the leaf space happens to be a manifold, then $A_{\mathcal F}/L_{\mathcal F}$ will be the space of sections of a vector bundle on $M/\mathcal F$ (the tangent bundle of $M/\mathcal F$), and thus so will be $A_{\mathcal F}$. Hence, in this case, the answer to your second question will be yes. 
5. Last question
I would guess the answer to your last question is no. This is because having $\{V(x)|V\in A_{\mathcal F}\}=T_xM$ for every $x\in U$ means that $D_{|U}=TM_{|U}$. Indeed,  if $V\in L_{\mathcal F}$ is non-zero at $x$, then we chose a function $f$ such that $(V\cdot f)(x)=1$ and we have 
$$
V'(x)=[V,fV'](x)-f(x)[V,V'](x)\in D_x
$$
for every vector field $V'$. Hence $D_{|U}=TM_{|U}$. 
Thus, if $U$ is moreover dense ($U$ being non-empty and $M$ connected is actually sufficient), then you get that $D=TM$. 
