Denote by $S$ your finite collection of $N$ points in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$.   Here is how you can recover $S$  from the knowledge of its  images via a finite collections of linear maps of rank $<n$.

Pick a  finite collection $\newcommand{\eL}{\mathscr{L}}$ $\eL$ of  linear maps  $\bR^n\to\bR$ in general position, i.e.,  any $n$ of them are linearly independent. Denote by $\nu$ the cardinality of $\eL$. The number $\nu$ is $> n$ and will be specified later.  For any collection $C\subset \eL$   we  obtain a  linear map

$$L_C:\bR^n\to\bR^C. $$

There are $\binom{\nu}{n-1}$ subsets $C\subset \eL$ of cardinality $n-1$. If $C$ is such a collection, then the linear map $L_C:\bR^n\to\bR^{n-1}$ is surjective and it has a one-dimensional kernel.  

**Suppose we know $L_C(S)$ for any collection $C\subset \eL$ of cardinality** $n-1$. Assume $\nu$ is large enough so that

$$\binom{\nu}{n-1}>\binom{N}{2}. $$

Since the $N$ points in $S$  determine  at most $\binom{N}{2}$ lines, we deduce that at least one of the linear maps $L_C$, $\# C=n-1$ restricts to an injective map $S\to \bR^C$.  In particular we deduce that

$$ N=\# S= \max_{\# C=n-1} L_C(S). $$

Choose $C_0\subset \eL$ such that $\# C_0=n-1$ and $\# L_{C_0}(S)=\# S=N$.  Without loss of generality we can assume that  $L_{C_0}$ is the  projection

$$P_0:\bR^n\to \bR^{n-1},\;\;(x_1,\dotsc,x_n)\mapsto (x_1,\dotsc, x_{n-1}). $$

For each point $s\in S$ we set $s':=P_0(s)$. Now we have complete knowledge of the  set

$$ S'=\bigl\lbrace\; s';\;\;s\in S\;\bigr\}=P_0(S). $$

The set $S'\subset \in\bR^{n-1}$ has the same cardinality as  $S$.  Moreover any point $s'\in S'$ determines a vertical line, i.e., a line parallel with $\ker P_0$, 


$$ \ell_{s'}=s'+\ker P_0=\bigl\{\; (s', t)\in\bR^n;\;\;t\in\bR\;\bigr\}. $$

We now have determined $N$ vertical lines and each one of them  contains exactly one point in $S$.     

**Suppose that we know $L(S)\subset \bR$ for any $L\in\eL$.**

Choose a linear functional $L\in \eL\setminus C_0$.  The set $L(S)$  has $m\leq N$ elements $r_1<\cdots <r_m$.   We obtain  $m$-hyperplanes

$$H_j(L)=\{ L(x)=r_j\},\;\;j=1,\dotsc, m, $$

and a set  $X(S,L)$ consisting of $Nm$ points

$$ H_j(L)\cap \ell_{s'},\;\;j=1,\dotsc, m,\;\;s'\in S'.  $$

Clearly $S\subset  X(S,L)$.  Thus $S$ can only be one of the $\binom{Nm}{N}$ subsets of $X$ of cardinality $Nm$.   Doing this with any $L\in \eL\setminus C_0$  we  deduce


$$ S\subset \bigcap_{L\in\eL\setminus C_0} X(S,L). $$


It seems plausible that if $\eL$ is large enough then

$$ \bigcap_{L\in\eL\setminus C_0} X(S,L)=S.  $$


Fix a linear map $L_0\in \eL\setminus C_0$ and set $X_0=X(S, L_0)$. We know that 

$$ S\subset X_0,\;\; \# X_0\leq N^2. $$

Suppose that $\nu$ is large enough so that

$$\binom{\nu}{n-1}>\binom{N^2}{2} +2.  $$

We can  then find  a collection $C_1\subset \eL$ of cardinality $n-1$ such that $C_1\neq C_0$ and $L_{C_1}$ and the restriction of $L_{C_1}$ to $X_0$ is injective.   We know know exactly   $L_{C_1}(X_0)$ and $S_1:=L_{C_1}(S)\subset L_{C_1}(X_0)$.  Note that $\# S_1=\# S=N$.

 For each point $s_1\in S_1$ we get a line $\ell_{s_1}= L_{C_1}^{-1}(s_1)$.    Let us observe that  each line $\ell_{s_1}$ intersects  exactly one of the lines  $\ell_{s'}$ because  

$$\ell_{s_1}\cap\ell_{s'}\subset X_0 $$

and the restriction of $L_{C_1}$ to $X_0$ is one-to-one.

> To conclude, if $\eL\subset \in {\rm Hom}\;(\bR^n,\bR)$ is a finite
> collection in general position whose cardinality  $\nu$ satisfies 
> 
> $$\binom{\nu}{n-1}>\binom{N^2}{2}+2, $$
> 
> and we know   $L_C(S)$ for any collection $C\subset \eL$ of
> cardinality $1$ or $n-1$, then we can  completely recover  $S$.