Is it true to say that the total space of the tautological line bundle over $\mathbb{R}P^{n}$ is a non orientable manifold?
Perhaps the question can be indirectly related to the following question:
The blow up of of the plane and the Moebius band
As another question which is indirectly related to our question we ask:
What is an example of a non orientable manifold $M$ such that $TM$ is an orientable manifold?
The Whitney product formula for Stiefel-Whitney classes specializes to $w_1(V\oplus W)=w_1(V)+w_1(W)$, where $w_1$ is the first Stiefel-Whitney class.
If $V$ is a vector bundle over $M$ (a CW-complex), $w_1(V)=0$ if and only if $V$ is orientable as a vector bundle.
If $E$ is the total space of a smooth vector bundle over a manifold $M$, then $T(E)$ always splits as $p^*(T(M))\oplus p^*(V)$ where $p:E \to M$ is the projection.
In particular, for any manifold $M$, the vector bundle $T(TM)$ splits as $p^*(TM)\oplus p^*(TM)$ where $p$ is the projection $TM \to M$. By the Whitney product formula(as mentioned in an answer to a linked question), $w_1(T(TM))=2w_1(p^*(TM))=0$.
Let $\gamma_n$ be the total space of the canonical bundle over $\Bbb RP^n$. Now $T(\gamma_n) \cong p^*(\gamma_n)\oplus p^*(T(\Bbb RP^n)). $ A standard computation shows that $w_1(\gamma_n)= \alpha$ the unique nonzero class in $H^1(\Bbb RP^n,\Bbb Z_2$). It is also standard to check that $\Bbb RP^n$ is orientable if and only if $n$ is odd. Equivalently $w_1(T(\Bbb RP^n))=\alpha$ if and only if $n$ is even. By naturality of $w_1$, $w_1(T(\gamma_n))=p^*(\alpha) +0\ne 0$ when $n$ is odd and $w_1(T(\gamma_n))=p^*(\alpha) +p^*(\alpha)=0$ when $n$ is even (which gives the total space orientable iff $n$ is even as mentioned in the comments).
One can generalize this idea quite easily to show the total space of a smooth vector bundle $V$ over a smooth manifold $M$ is orientable if and only if $w_1(TM)=w_1(V)$.