What is the dimension of a tangent space? Let $U \subset \mathbb{R}^2$ be open and $f: U \to \mathbb{R}^3$ be an immersion (so that $U$ and $f(U)$ are diffeomorphic).
Let $p \in U$. Thus $f(p)$ is a point on the surface $f(U) \subset \mathbb{R}^3$.
Then which of the following is correct? 

1. $\dim T_p U = 2$.  
2. $\dim T_{f(p)} f(U) = 2$.  
3. $\dim T_{f(p)}\mathbb{R}^3 = 3$.  
4. $T_{f(p)}f(U) \subset T_{f(p)} \mathbb{R}^3$.  
5. The surface normal $N(p)$ is always an element of $T_{f(p)}\mathbb{R}^3$ but never an element of $T_{f(p)}f(U)$. 
6. If $v \in T_p U$, then $v$ is of the form $(p, (v_1, v_2))$.
7. If $v \in T_{f(p)} f(U)$ then $v$ is of the form $(f(p), (v_1, v_2, 0))$ and thus $T_p U \simeq T_{f(p)} f(U)$.  
8. If $w \in \left[T_{f(p)} \mathbb{R}^3 \setminus T_{f(p)}f(U)\right]$ then it can not be identified with a $v \in T_p U$. 
9. $d_p f: T_p U \to T_{f(p)}\mathbb{R^3}$ (in particular, it does not map inside of $T_{f(p)}f(U)$ which is 2, not 3, dimensional) can be identified with a $3 \times 2$ matrix (as opposed to a $2 \times 2$ matrix).

I am using this to try to learn about the shape operator.
 A: I will refer to your items by number in the order in which they do occur, so if you change the order it will become wrong.
Some of these look ill-posed to me. Especially the last one. While $d_pf$ can in fact be identified with a $3\times 2 $ matrix, it does map the source into $T_{f(p)} f(U)$ which can be identified with a subspace of $T_{f(p)} \mathbb{R}^3$ (so the remark starting with 'in particular' is incorrect). Depending on the representation chosen it can also be represented by a $2\times 2$ matrix (with respect to a choice of base vector fields on both $T_p(U)$ and $T_{f(p)} f(U)$, at least locally.
Also note that an immersion may have a self-intersecting image. You can only locally claim that it is diffeomorphism on the image. Assuming you know it is a global diffeomorphism some of your question are a bit easier to respond to (I will assume this is the case). 
Then 
1) - 6) are true, 7) is not, assuming you are using the standard Euclidean Coordinate system in the target space. (it can be of the form $(f(p), (v_1, v_2, v_3))$. 8) is correct again, if, by 'identify' you mean by means of $d_pf$.
9) I already discussed.
I would rephrase 6) as 'can be expressed in the form'
Regarding the follow up queustion in the comment: locally you can choose a unit normal vector field along the surface. With $f$ and the normal vector field you can define coordinates in the neighbourhood of $f(U)$ (you may have to make $U$ smaller for this to work) in which $f(U)$ is represented by $x_3=0$ -- you just use the signed normal distance to $f(U)$ as third coordinate and the image of the coordinates in $U\subset \mathbb{R}^2$, $(\bar x_1, \bar x_2)$ under $f$ as the first two coordinates $(x_1,x_3)$ in the image. That this is in fact a coordinate system follows from the implicit function theorem after some calculations (you may try to search for 'normal' coordinates).
