About concept: how to pull back a manifold function that is not injective? Given $F:M\rightarrow N$ that is not injective, I have a great confusion on  the pull-back map $F^*:T^*_{F(p)}N\rightarrow T^*_pM$.  For better communication, let's simply take $M=\mathbb{R}^2$, $N=\mathbb{R}$. Take $(x_1,x_2)\in M$, $F(x_1,x_2)=x_1\in N$. Let $f\in C^\infty(N)$.
(1) When we pull back $f^\prime(x)d x\in T_x^*N$, I expect we get $T^*_{(x_1,x_2)}M\ni f^\prime (x_1)dx_1+0dx_2=F^*(f^\prime(x)d x)$. Then excuse me, how do we determine the value of $x_2$? This is basically a question on how to deal with the inverse map when a map is not injective.
(2) Also the resource I am reading says "(a) $F^*$ is NOT a map
from $T^*N$ to $T^*M.$
(b) It is a map from $C^\infty(N;T^*N)$ to $C^\infty(M;T^*M)$ which is quite a
different thing."
How should I understand this remark, especially in the combination with the first question? (I don't think neither (a) or (b) is a map at this moment, and I think (b) somehow would imply (a).)
Little remark: Maybe the answer to my question is simply to take $x_2$ to be any/all $x_2\in \mathbb{R}$?
Been confused on this for a while, hope I can get some enlightenment from you, thanks!
 A: If $M$ and $N$ are two manifolds and $f : M \to N$ is smooth, any vector bundle $E$ on $N$ can be pulled-back on $M$ by the following construction:
$$
f^*E=\left\{(x,v) \in M\times E ~|~ v \in E_{f(x)} \right\}
$$
which basically says that we glue the vector space $E_{f(x)}$, the fiber of $E$ over $f(x)$, to $x$.
This way, if two points $x_1$ or $x_2$ in $M$ have the same image, then the fibers over $x_1$ and $x_2$ of $f^*E$ are exactly the same vector space. But this is not a confusing definition: for example, $M\times \mathbb{R}$ is a vector bundle over $M$ where all the fibers are exactly the same.
Now, if $s$ is a section of $E$ as a vector bundle over $N$, one can pull back this section as a section $f^*s$ over $M$ with the following constuction:
$$
(f^*s) : x \mapsto s_{f(x)}
$$
that is the value of the section $f^*s$ at a point $x\in M$ is the value of a the section $s$ at the point $f(x)\in N$. Again, if $f(x_1)=f(x_2)$, the fibers of $f^*E$ over $x_1$ and $x_2$ are the same, and the value of the pulled-back section are the same at $x_1$ and $x_2$ as seen in the fibers.
If, now, $T$ is a tensor over $N$, then one can ask how to evaluate such a tensor. This construction cannot allow you to talk about the pullback of a vector field on $N$ as a vector field on $M$! Indeed, if $X$ is a vector field on $N$, then $f^*X$ is a section of $f^*(TN)$ which is not $TM$.
But if $T$ is a covariant vector field, there is no problem here. For example, if $\alpha$ is a $1$-form on $N$, then one can decide that $f^{\alpha}$ is a $1$-form on $M$ by saying
$$
(f^*\alpha)_x V_x = \alpha_{f(x)}(\mathrm{d}f(x)V_x)
$$
And similarly, for a $p$-covariant tensor,
$$
(f^*T)(V_1,\ldots,V_p)_x = T_{f(x)}\left(\mathrm{d}f(x){V_1}_x,\ldots,\mathrm{d}f(x){V_p}_{x}\right)
$$
This defined a $p$-covariant tensor on $M$.
Briefly, any smooth map $f : M \to N$ allows you to pullback any vector bundle and any section of this vector bundle. But these pulled-back bundles are not generally usual vector bundles as the tensor bundles.
Any covariant tensor bundle $(T^*N)^{\otimes p}$ can pulled back as a canonical sub-bundle $f^*(T^*N)^{\otimes p}\subset (T^*M)^{\otimes p}$ using $\mathrm{d}f$ (thanks to the covariance). But $TN$ cannot be canonically pulled-back as a subbundle of $TM$, neither are $TM^{\otimes q}$. One can outpass this if $\mathrm{d}f$ is an isomorphism between $T_xM$ and $T_{f(x)}N$ for all $x$.
In you example, where $f(x,y) = x$ from $\mathbb{R}^2 \to \mathbb{R}$, if $g$ is a smooth function on $\mathbb{R}$, then $\mathrm{d}g = g' \mathrm{d}t$ is a $1$-form on $\mathbb{R}$, and $f^* (\mathrm{d}g)$ is the one form on $\mathbb{R}^2$ defined by the folliwing, if $X=(X_1,X_2)$ is a vector field on $\mathbb{R}^2$
$$
f^*(\mathrm{d}g) = \mathrm{d}g(f(x,y)) \mathrm{d}f(X_1,X_1) = \mathrm{d}g(x)X_1
$$
