I think I don´t have trouble with the first part, because for the antipodal map $f:\mathbb{S}^2\rightarrow \mathbb{S}^2, \ \ p \mapsto -p$, and any differentiable atlas for the sphere $\mathcal{A}=\{(U_\alpha, \phi_{\alpha})\}$, then $\mathcal{B}=\{(f^{-1}(U_\alpha), \phi_{\alpha}\circ f )\}$ is a differentiable atlas for the sphere that is compatible with $\mathcal{A}$, then the quotient space is a smooth 3-manifold and the projection $$ \begin{array}{rcl} \pi: \mathbb{S}^2\times [0,1] & \longrightarrow & M \\ (p,t) & \longmapsto & [p,t] \end{array} $$ is a smooth map. Moreover, for each $t$, the fiber $\{\pi^{-1}[p,t]\ \vert \ p\in \mathbb{S}^2\}$ is diffeomorphic to $\mathbb{S}^2$.
Some ideas for the orientability?
Assume that $M$ is oriented and admits a volume form $\omega$. We will show that it implies that $f$ is orientation preserving, which is false. Let $I = \left]-\frac{1}{2},1\right[$ and, $$ \psi : \left\{\begin{array}{rcl} \mathbb{S}^2 \times I & \rightarrow & M \\ (p,t),\ t \leqslant 0 & \mapsto & [f(p),t + 1] \\ (p,t),\ t \geqslant 0 & \mapsto & [p,t] \end{array}\right. $$ $\psi_I$ is a local diffeomorphism. Therefore, $\psi^*\omega$ is a volume form on $\mathbb{S}^2 \times I$ and defines an orientation on it. Now, let $J = \left]-\frac{1}{2},\frac{1}{2}\right[ \subset I$ and $K = ]0,1[ \subset I$, let $\sigma : (p,t) \mapsto \left(p,t + \frac{1}{2}\right)$ from $\mathbb{S}^2 \times J$ to $\mathbb{S}^2 \times K$ and $\iota_J : \mathbb{S}^2 \times J \hookrightarrow \mathbb{S}^2 \times I$ and $\iota_K : \mathbb{S}^2 \times K \hookrightarrow \mathbb{S}^2 \times I$ the inclusions. Notice that $\psi_{|\mathbb{S}^2 \times J}$ and $\psi_{|\mathbb{S}^2 \times K}$ both are isomorphisms onto their images. Moreover, $\psi_{|\mathbb{S}^2 \times K} \circ \sigma$ is homotopic to $\psi_{|\mathbb{S}^2 \times J}$ using the homotopy, $$ H : ((p,t),s) \mapsto \psi\left(p,t + \frac{s}{2}\right) $$ In particular, $\psi_{|\mathbb{S}^2 \times J}^{-1} \circ \psi_{|\mathbb{S}^2 \times K} \circ \sigma$ preserves the orientation on $\mathbb{S}^2 \times J$. And for all $(p,t) \in \mathbb{S}^2 \times J$, \begin{align*} \left(\psi_{|\mathbb{S}^2 \times J}^{-1} \circ \psi_{|\mathbb{S}^2 \times K} \circ \sigma\right)(p,t) & = \left(\psi_{|\mathbb{S}^2 \times J}^{-1} \circ \psi_{|\mathbb{S}^2 \times K}\right)(p,t + 1/2)\\ & = \psi_{|\mathbb{S}^2 \times J}^{-1}([p,t + 1/2])\\ & = \psi_{|\mathbb{S}^2 \times J}^{-1}([f(f^{-1}(p)),t - 1/2 + 1])\\ & = (f^{-1}(p),t - 1/2). \end{align*} This map preserve the orientation, which implies that $f^{-1}$ preserves the orientation on $\mathbb{S}^2$, which implies that $f$ preserves it too, which is a contradiction.
Notice that the same proof works if you replace $\mathbb{S}^2$ by any orientable manifold $N$ (even topological ones, but in this case, you have to replace volume forms by generators of $H^{\dim(N)}(N)$), and any $f : N \rightarrow N$ orientation reversing homeomorphism.