I think that the OP is asking a more specific question than whether or not a surface has a connection that is not metric or not torsion free. It seems that the OP is assuming that the surface $M$ comes equipped with an immersion $\mathbf{r}:M\to\mathbb{E}^3$ into (oriented) Euclidean $3$-space and is asking whether, *using the data of the immersion $\mathbf{r}$*, it is possible to define, in a canonical way, a connection that has torsion and/or is not metric compatible.

His question includes the argument that the usual induced connection associated to a given $\mathbf{r}$ discussed in all curves-and-surfaces books is both compatible with the induced metric and is torsion-free.

Now, it's true that the only canonical connection induced by $\mathbf{r}$ that uses at most *second-order* information from $\mathbf{x}$ at a point is the Levi-Civita connection. However, there are other canonical connections definable using $\mathbf{r}$ that use higher order information, and these need be neither torsion-free nor compatible with any metric (let alone the induced metric), at least for the general immersion. (Obviously, any canonical formula using higher order information will just produce the Levi-Civita connection when applied to an immersion whose image is either a plane or a sphere.)

**Example**: Given an immersion $\mathbf{x}:M\to\mathbb{E}^3$, there is an associated mean curvature function $H$ that, unfortunately, depends on a choice of orientation of the surface $M$; it switches sign if one reverses the orientation of $M$ (always, assuming, of course, that the target space $\mathbb{E}^3$ is oriented). However, the $1$-form $\eta = \ast dH$ is independent of a choice of orientation of the surface, since both $H$ and $\ast$ reverse sign when one reverses orientation. Let $\nabla$ be the Levi-Civita connection on $M$ associated to the metric induced on $M$ by the immersion $\mathbf{x}$, and define a second connection $\tilde\nabla$ on $M$ by the formula
$$
\tilde\nabla_XY = \nabla_XY + \eta(X)Y
$$
Then $\tilde\nabla$ is a connection canonically associated to $\mathbf{x}$ (whose local formula depends on third order derivatives of $\mathbf{x}$). One computes (using the fact that the torsion of $\nabla$ vanishes) that
$$
T^{\tilde\nabla}(X,Y) = \tilde\nabla_XY - \tilde\nabla_YX - [X,Y] = \eta(X)Y - \eta(Y)X,
$$
so the torsion of $\tilde\nabla$ vanishes if and only if $\eta=0$, i.e., $H$ is locally constant.

Meanwhile, it is easy to compute that the curvatures of the two connections are related by
$$
R^{\tilde\nabla}(X,Y)Z = R^{\nabla}(X,Y)Z + d\eta(X,Y)\ Z,
$$
so $\tilde\nabla$ does not even have a parallel $2$-form, let alone a parallel metric, unless $d\eta=0$, i.e., unless $H$ is (locally) a harmonic function on the surface.

Thus, in general, $\tilde\nabla$ is neither torsion-free nor metric compatible.

`$\mathbf{r}=\theta^{\alpha}\mathbf{e}_\alpha$`

? $\endgroup$ – Ayan Jun 11 '13 at 12:25existmany such connections, just choose arbitrary functions $\Gamma^\alpha_{\beta\gamma}(\theta^1,\theta^2)$ and define a connection by $$ \nabla_{\mathbf{e}_\beta}\mathbf{e}_\gamma = \Gamma^\alpha_{\beta\gamma}\ \mathbf{e}_\alpha\ . $$ However, I grant you that this method makes arbitrary choices that are not forced by the given immersion. If you want to make it depend canonically on the given immersion, that's another matter. See my answer below. $\endgroup$ – Robert Bryant Jun 11 '13 at 12:42