Dimensions of velocity vector in differential geometry If we have a velocity vector written in, say, cartesian coordinates:
$\mathbf V$ = $\dot{x}$ i $+$ $\dot{y}$ j
Note that Dim($\dot{x}$) = Dim($\dot{y}$) = $LT^{-1}$ which are the dimensions of speed and Dim(i) $=$ Dim (j) $= 1$ (i.e. dimensionless since they are unit vectors). Thus we have the totally expected result that the dimensions of velocity are $LT^{-1}$. Moreover, we have a straightforward interpretation of the components of this vector as being the speed of the (say) particle in the x and y directions respectively.
Going to the differential geometry setting we write our vector as:
$\mathbf V =A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y}$
I write $A$ and $B$ because I no longer know how to interpret the components, this is because the dimensions of the partial derivative operators are (seemingly) $L^{-1}$ as opposed to the dimensionless basis vectors we had above. But this means that $A$ and $B$ must have dimensions of $L^{2}T^{-1}$ for the velocity vector as a whole to have the correct dimensions. So what is going on here? Can we no longer interpret the components of this vector as we did above? Also please clarify the issue for the two cases where our manifold has and doesn't have a metric because I suspect if you do have a metric you can somehow normalize the vectors (for instance defining a new basis vector which equals $\frac{\partial}{\partial x}$ divided by $\sqrt{(g_{11}})$ would take care of the issue I think but i'm not entirely sure).
 A: Given a curve $\gamma: \mathbb R \rightarrow \mathcal M$ where $\mathcal M$ is the manifold under consideration (in your case, presumably $\mathbb R^2$), we define the tangent vector to $\gamma$ at the point $p=\gamma(0)$ to be a map $\mathcal V_{\gamma,p}$ which eats smooth functions $f:\mathcal M\rightarrow \mathbb R$ and spits out the following number:
$$\mathcal V_{\gamma,p} (f) = (f\circ \gamma)'(0)$$
In other words, a tangent vector is a directional derivative operator associated to some curve at some point.  It eats a function and spits out the rate at which that function changes along the associated curve.  Therefore, if we interpret the curve parameter as time, then tangent vectors have dimension $\mathrm T^{-1}$.
In a coordinate chart where $\gamma(t) = \big(x(t),y(t)\big)$, we have
$$V_{\gamma,p}(f) = \frac{\partial f}{\partial x}x'(0) + \frac{\partial f}{\partial y}y'(0) \iff V_{\gamma,p} = x'(0) \frac{\partial }{\partial x} + y'(0) \frac{\partial }{\partial y}$$
If the coordinates $(x,y)$ have dimensions of $\mathrm L$, then the components of the vector in this basis have dimensions of $\mathrm{LT}^{-1}$, and the basis vectors themselves have dimensions of $\mathrm L^{-1}$.

Your mistake lies in the following line:

But this means that A and B must have dimensions of $L^2 T^{-1}$ for the velocity vector as a whole to have the correct dimensions.

You are assuming that the velocity vector has dimensions of $\mathrm L \mathrm T^{-1}$, but in fact the correct dimensions of a directional derivative operator are $\mathrm T^{-1}$ (again, assuming that the curve parameter has dimensions of $\mathrm T$).

Also please clarify the issue for the two cases where our manifold has and doesn't have a metric

The presence or absence of a metric has no bearing on this issue.  You can always define new dimensionless coordinates $\tilde x = x/x_0$ and $\tilde y = y/y_0$.  However, the dimensions of a tangent vector are $\mathrm T^{-1}$ regardless of your coordinates (which should be obvious, since the definition of the tangent vector is coordinate-free), and so making this redefinition will yield velocity components with dimensions of $\mathrm T^{-1}$.  I suspect this is not what you want.
