Consider the space of non zero real homogeneous degree $2$ polynomials in three variables upto scaling. This space is $\mathbb{R} \mathbb{P}^5$. The zero set 
of such a polynomial gives a real curve in $\mathbb{R} \mathbb{P}^2$. Let me 
define the space $X$ to be the space of real curves $[f]$ and a marked point 
$p$, such that the curve has a singularity at $p$, i.e. 
$$ X := \{ ([f], p) \in \mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2: 
f(p) =0, ~~\nabla f|_p =0 \}. $$ 

$\textbf{Question $1$:}$ Is $X$ a smooth manifold of the expected dimension (four)?

$\textbf{Question $2$:}$ If yes, is $X$ orientable or non orientable?   

$\textbf{Comments:}$ I have two arguments giving me contradictory answers. 
First of all, note that (assuming $X$ is a manifold), 
the normal bundle to $X$ in 
$\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2$ is given by 
$$ N_X:= \gamma_{5}^* \otimes \gamma_2^{* 2} \oplus \gamma_5^* \otimes T^*\mathbb{R P}^2 \otimes \gamma_2^* \big{|}_X, $$ 
where $\gamma_n$ is the tautological line bundle over $\mathbb{RP}^n$. 
(I can justify this if someone is not convinced). 

Note that 
$$ TX \approx 
\frac{T (\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2)}{N_X}. $$
It is now easy to see that the first stiefel whitney class of the 
tangent bundle of $X$ is zero, i.e. $w_1(TX) =0$. Hence $X$ is orientable. 

On the other hand it seems to me that  $~\pi: X \rightarrow \mathbb{RP}^2$ 
is a fibre bundle, with fibres $\mathbb{RP}^2$. One can show (using spectral sequences) that any $\mathbb{RP}^2$ bundle over $\mathbb{RP}^2$ is non orientable. 

$\textbf{Proof of why $X$ is a manifold:} $ Let us take $\mathbb{R}^6$ 
to be the space of real polynomials in two variables of degree at most $2$. Hence we can write such an element $\rho \in \mathbb{R}^6$ as 
$$ \rho(x,y) = \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2. $$

Now consider the map 
$$ \psi : \mathbb{R}^6 \times \mathbb{R}^2 \rightarrow \mathbb{R}^3 $$ 
given by 
$$ \psi(\rho; x,y) := \Big( \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2, \\
~\rho_{10}  + \rho_{20}x + \rho_{11} y, \\ 
~\rho_{01}  + \rho_{11}x + \rho_{02}y \Big). $$ 
It is easy to see that if $\psi(\rho,0,0) =0$, then the Jacobian matrix 
of $\psi(\rho,x,y)$ at $(x,y)=(0,0)$ has full rank. To see why, take the 
partial derivative of $\psi$ with respect to $\rho_{00}$, $\rho_{10}$ 
and $\rho_{01}$ and plug in $(x,y) = (0,0)$. That gives us a $3\times 3$ 
identity matrix. Hence $\psi$ is transverse to the zero 
set (it is easy to see taking $(x,y) = (0,0)$ was without any loss of generality).
Hence, by Implicit Function Theorem, $\psi^{-1}(0,0,0)$ 
is a smooth manifold (even around double lines). 

Its now easy to see that the space $X$ defined will also be a manifold; 
by writing the evaluation map and the vertical derivative 
in a coordinate chart and trivialization, it reduces to the above calculation.