The retraction (not deformation retraction) of a torus onto only one circle can be seen geometrically by considering a torus which has the two circles intersecting $\mathbb{R}^2$ being a circle with radius $1$ and one with radius $2$. Take the projection $\pi: \mathbb{R}^3 \to \mathbb{R}^2$ on the first two coordinates, and the map $p:\mathbb{R}^2 -\{0\} \to S^1$ given by $p(x)=x/\Vert x\Vert$. Then $p \circ \pi $ is a retraction onto the circle. One more elementary way is just to take the projection $\pi: S^1 \times S^1 \to S^1$ on the first coordinate.
The wedge of two circles can't be a retract of the torus, since its fundamental group is nonabelian, and hence can't be a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$.
Furthermore, we have that $(T^2,S^1)$ forms a "good pair". Note that $T^2/S^1 \sim S^2 \vee S^1$ (homotopy equivalent), which gives everything you want to calculate the homology of $T^2$.
For the $S_g$ part, consider the following image. I'll leave to you to understand what is the retraction.
(the case of $g>1$ is just collapsing everything beyond a certain height to a point).
For the second question, a similar argument used to prove that the wedge of two circles isn't a retract of the torus can be used to show that the mobius strip can't be a retract of $\mathbb{R}P^2$ as you suggest, since $\mathbb{Z}$ is not a subgroup of $\mathbb{Z}/2\mathbb{Z}$. Now, since $\mathbb{R}P^2$ is a mobius strip attached with a disk, if you remove that disk to make the connected sum, you get two mobius strips which you can simply project one onto the other (similarly to what happened above) to yield a retraction from $N_1$ to a mobius strip. The case of $g>2$ is similar.
Since you're a beginner at Algebraic Topology this topic might be a bit difficult (even as a grad student in algebraic topology I always found twisted coefficients to be difficult/hard to work with explicitly). Kirk and Davis' book "Lecture Notes in Algebraic Topology" has a chapter covering homology with local (aka twisted) coefficients starting on page 97. Here is how I managed to understand the relation between $H^*(M;\mathbb{Q}^\omega)$ and $H^*(M;\mathbb{Q})$:
Claim: there is a twisted version of Poincaré Duality with rational coefficients. Kirk and Davis state it for $\mathbb{Z}$ on page 104, and then below state that there is a generalization for any $\mathbb{Z}[\pi_1(M)]$ module $A$, and we can take $A=\mathbb{Q}$ where $\alpha\in\pi_1(M)$ acts by $-1$ if the tangent bundle is non-orientable over a loop representing $\alpha$, and acts trivially otherwise. In particular, if $M$ is a closed, connected manifold of dimension $n$ and $\mathbb{Q}^\omega$ is the coefficient system twisted by the orientation character then there is a twisted fundamental class $[M] \in H_n(M; \mathbb{Q}^\omega) \cong \mathbb{Q}$ inducing isomorphisms
$$ - \cap [M]\colon H^k(M;\mathbb{Q}) \cong H_{n-k}(M;\mathbb{Q}^\omega) $$
$$ - \cap [M]\colon H^k(M;\mathbb{Q}^\omega) \cong H_{n-k}(M;\mathbb{Q})$$
Remarkably, if $M$ is non-orientable this implies that $H^0(M;\mathbb{Q}^\omega) \cong H_0(M;\mathbb{Q}^\omega) \cong 0$ (as Captain Lama points out in their comment, the intuition behind this is that the orientation sheaf has no global section when the manifold is non-orientable).
Now, since $\mathbb{Q}$ is injective as a $\mathbb{Z}$-module, the Universal Coefficient Theorem gives us an isomorphism $\kappa\colon H^k(M;\mathbb{Q})\cong Hom(H_k(M;\mathbb{Z}), \mathbb{Q})$. But $$Hom(H_k(M;\mathbb{Z}), \mathbb{Q}) \cong Hom(H_k(M;\mathbb{Z})\otimes \mathbb{Q}, \mathbb{Q}) \cong H_k(M; \mathbb{Z})\otimes\mathbb{Q}\cong H_k(M;\mathbb{Q})$$
Here the first isomorphism is because a homomorphism to $\mathbb{Q}$ will kill any torsion so we can ignore it by tensoring with $\mathbb{Q}$, the second isomorphism is because they are vector spaces, and the third isomorphism is by the Universal Coefficient Theorem for Homology and uses the fact that $\mathbb{Q}$ is torsion-free.
Combined together we actually find
$$H^k(M;\mathbb{Q}^\omega) \cong H_{n-k}(M;\mathbb{Q})\cong H^{n-k}(M;\mathbb{Q}) $$
Best Answer
Yes, $H^{n-1}(M-\{p\};\mathbb{Q}) \neq 0$ for any closed non-orientable manifold of dimension $n$. Long-story-short it's because $H^n(M)=0$ in this case so $H^{n-1}(M-\{p\})$ surjects onto the coefficient group.
Just for exposition I wanted to spell out the arguments to a few of your assertions. Consider the long exact sequence of the pair:
$$\dots \to H^{k}(M, M-\{p\}) \to H^k(M) \to H^k(M-\{ p\})$$$$ \to H^{k+1}(M, M-\{p\}) \to H^{k+1}(M)\to \dots $$
This sequence exists for all coefficients $R$, in particular $\mathbb{Q}$.
Since $M$ is a manifold of dimension $n$, the group $H^k(M, M-\{p\})$ is $R$ if $k = n$ and $0$ otherwise. It follows that $$H^k(M) \cong H^k(M-\{p\})\text{ if $k < n-1$}$$
and there is an exact sequence
$$0 \to H^{n-1}(M) \to H^{n-1}(M-\{ p\}) \to R \to H^n(M) \to 0$$
where $H^n(M-\{p\}) = 0$ because it is an open manifold.
If $M$ is orientable then the last map is an isomorphism so $H^{n-1}(M) \cong H^{n-1}(M-\{p\})$. Otherwise $M$ is non-orientable, so if $R$ is such that $H^n(M) =0$ (for example $R=\mathbb{Z}, \mathbb{Q}$) then you get a short exact sequence with $H^{n-1}(M-\{p\})$ in the middle: if $R=\mathbb{Q}$ that means $H^{n-1}(M-\{p\})$ is a rational vector space of dimension $dim(H^{n-1}(M)) +1>0$. In fact as long as $H^n(M)=0$ and $R\neq 0$ we must have $H^{n-1}(M-\{p\};R)\neq 0$ because it surjects onto $R$.