What is the image of $x^{\rm T}Qx\le 1$ under a linear map $x \mapsto Cx$?

By changing the orthonormal bases in $\mathbb R^n$ and $\mathbb R^m$, we may assume that $C=\pmatrix{D&0}$ for some nonsingular matrix $D$. Let $$ Q=\pmatrix{X&Y^T\\ Y&Z}, \ M=\pmatrix{I_m&0\\ -Z^+Y&I_{n-m}} \ \text{ and } \ x=\pmatrix{u\\ v}.\tag{1} $$ As $Q$ is positive semidefinite, the range of $Y$ must lie inside the range of $Z$, i.e. $Y=ZW$ for some matrix $W$. It follows that $Y-ZZ^+Y=(Z-ZZ^+Z)W=0$. Hence $$ M^TQM=\pmatrix{X-Y^TZ^+Y&0\\ 0&Z} \ \text{ and } \ M^{-1}x=\pmatrix{u\\ v+Z^+Yu}. $$ Since $x^TQx=(M^{-1}x)^T(M^TQM)(M^{-1}x)$, we obtain $$ x^TQx=u^T(X-Y^TZ^+Y)u+(v+Z^+Yu)^TZ(v+Z^+Yu).\tag{2} $$ $M^TQM$ must be positive semidefinite because it is congruent to $Q$. Thus $X-Y^TZ^+Y$ and $Z$ are also PSD. Now define $$ R:=(D^{-1})^T(X-Y^TZ^+Y)D^{-1}\ \text{ and }\ y:=Cx=Du.\tag{3} $$ If $y=Cx$ for some $x$ with $x^TQx\le1$, $(3)$ shows that $u=D^{-1}y$ is uniquely determined by $y$ and $(2)$ shows that $u^T(X-Y^TZ^+Y)u\le1$. Yet, $u^T(X-Y^TZ^+Y)u$ is precisely $y^TRy$. Therefore $y^TRy\le1$. Conversely, if $y^TRy\le1$, put $u=D^{-1}y$ and $v=-Z^+Yu$ in $(1)$. Then $(2)$ shows that $x^TQx\le1$. Therefore $$ \{y:y^TRy\le1\}=\{y:y=Cx \text{ for some $x$ with $x^TQx\le1$}\}. $$ It remains to express $R$ in terms of $Q$ and $C$. Let $P=\pmatrix{0&0\\ 0&I_{n-m}}=I-C^+C$. Then \begin{align} R&=\pmatrix{(D^{-1})^T&0}\left[\pmatrix{X&Y^T\\ Y&Z}-\pmatrix{0&Y^T\\ 0&Z}\pmatrix{0&0\\ 0&Z^+}\pmatrix{0&0\\ Y&Z}\right]\pmatrix{D^{-1}\\ 0}\\ &=(C^+)^T\left[Q-QP(PQP)^+PQ\right]C^+\\ &=(C^+)^TQ^{1/2}\left[I-A(A^TA)^+A^T\right]Q^{1/2}C^+\quad(A=Q^{1/2}P)\\ &=(C^+)^TQ^{1/2}(I-AA^+)Q^{1/2}C^+\\ &=(C^+)^TQ^{1/2}\left[I-\left(Q^{1/2}(I-C^+C)\right)\left(Q^{1/2}(I-C^+C)\right)^+\right]Q^{1/2}C^+.\tag{4} \end{align} Now $(4)$ our basis-independent formula for $R$. It has the following geometric interpretation. Basically, we want to find a semi-inner product $\langle\cdot,\cdot\rangle_{\mathbb R^m}$ such that $\langle y,y\rangle_{\mathbb R^m}\le1$ if and only if $y=Cx$ for some $x$ with $x^TQx\le1$. Since $x^TQx=(Q^{1/2}x,\,Q^{1/2}x)$, where $(\cdot,\cdot)$ denotes the standard inner product on $\mathbb R^n$, an obvious strategy is to map $y\in\mathbb R^m$ to the vector $x=C^+y\in\mathbb R^n$ and define $\langle y,y\rangle_{\mathbb R^m}$ as $\langle Q^{1/2}x,Q^{1/2}x\rangle_{\mathbb R^n}$ for some appropriate semi-inner product defined on $\mathbb R^n$. Since the solution set for the equation $Cx=y$ is given $C^+y+\ker(C)$, we want $(\cdot,\cdot)$ to be zero on $Q^{1/2}\ker(C)$. A natural choice is therefore to define $\langle u,v\rangle_{\mathbb R^n}$ as $(Pu,Pv)$, where $P$ is the orthogonal projection onto $\left(Q^{1/2}\ker(C)\right)^\perp$. In short, we define $$ \langle y,y\rangle_{\mathbb R^m}=(PQ^{1/2}C^+y,PQ^{1/2}C^+y).\tag{5} $$ It is straightforward to verify that this semi-inner product does work. Suppose $x^TQx\le1$ and $y=Cx$. Since $PQ^{1/2}(I-C^+C)=0$, we have $PQ^{1/2}=PQ^{1/2}C^+C$. Therefore \begin{aligned} \langle y,y\rangle_{\mathbb R^m} &=(PQ^{1/2}C^+y,PQ^{1/2}C^+y)\\ &=(PQ^{1/2}C^+Cx,PQ^{1/2}C^+Cx)\\ &=(PQ^{1/2}x,PQ^{1/2}x)\\ &\le(Q^{1/2}x,Q^{1/2}x)\\ &=x^TQx\\ &\le1. \end{aligned} Conversely, suppose $\langle y,y\rangle_{\mathbb R^m}=(PQ^{1/2}C^+y,PQ^{1/2}C^+y)\le1$. Let $Q^{1/2}(I-CC^+)z$ be the orthogonal projection of $Q^{1/2}C^+y$ onto $Q^{1/2}\ker(C)$ and let $x=C^+y-(I-CC^+)z$. Then $y=Cx$. Moreover, as $Q^{1/2}x=Q^{1/2}C^+y-Q^{1/2}(I-CC^+)z\in\left(Q^{1/2}\ker(C)\right)^\perp$, we have \begin{aligned} 1&\ge\langle y,y\rangle_{\mathbb R^m}\\ &=(PQ^{1/2}C^+y,PQ^{1/2}C^+y)\\ &=(PQ^{1/2}x,PQ^{1/2}x)\\ &=(Q^{1/2}x,Q^{1/2}x)\quad\text{because }Q^{1/2}x\in\left(Q^{1/2}\ker(C)\right)^\perp\\ &=x^TQx. \end{aligned} Now, if we write $(5)$ in matrix form, we get $(4)$.