Yes, the Jacobian relates the joint velocities to end-effector velocity through this equation:
$$
\mathbf{v}_e = \mathbf{J}(\mathbf{q}) \dot{\mathbf{q}}
$$
Where $\mathbf{q}$ is the joint angles, $\dot{\mathbf{q}}$ is the joint velocities, and $\mathbf{v}_e$ is the end-effector velocity. As you can see, the Jacobian, $\mathbf{J}$, is configuration dependant. So plug in some joint angles and velocities then you will get the velocity of the end-effector.
To get the Cartesian position of the end-effector, given the joint angles, you use the direct kinematics function also called forward kinematics. There are various methods to do this. Geometric analysis if your arm is simple enough e.g. a planar 2 link arm. Product-of-exponentials is another method. But the Denavit-Hartenberg method is probably the most widely used. I am not going to go into the details of this here. But basically you will get a transformation matrix for each joint: $\mathbf{A}_i^{i-1}(q_i)$. That when you plug in the joint angle, you get the pose of joint $i$ relative to joint $i-1$. These can be combined in a recursive fashion to get the pose of the end-effector relative to the base of the arm:
$$
\mathbf{T}_n^0(\mathbf{q})=\mathbf{A}_1^{0}(q_1) \mathbf{A}_2^{1}(q_2) ... \mathbf{A}_n^{n-1}(q_n)
$$
Note that you can differentiate $\mathbf{T}_n^0(\mathbf{q})$ to get the analytical Jacobian. But people typically use the geometric Jacobian which is not as hard to compute.
Now to compute the required joint velocities to achieve a desired end-effector velocity, you must invert the Jacobian. But this only works if the number of DOFs equals the number of dimensions of your space:
$$
\dot{\mathbf{q}} = \mathbf{J}^{-1}(\mathbf{q}) \mathbf{v}_e
$$
(Note that there are some arm configurations (such as singularities) where the Jacobian will not be invertable.) If you have more DOFs, you are under-constrained. (i.e. there is more than one solution). Typically, people use the right pseudo-inverse of the Jacobian which locally minimizes the norm of joint velocities.
$$
\dot{\mathbf{q}}=\mathbf{J}^{\dagger}\mathbf{v}_{e}
$$
where:
$$
\mathbf{J}^{\dagger} = \mathbf{J}^T(\mathbf{J}\mathbf{J}^T)^{-1}
$$
Note that J is still dependant on q, but (q) is dropped for clarity.
