How does a tangential force cause a linear acceleration? Consider a rigid sphere in empty space of mass m and radius r. If a tangential force(torque) F (or any force not along the line of the centre of mass) is applied, it is my knowledge the sphere will translate and rotate. its COM will translate with an acceleration F/m, and it will rotate with an angular acceleration of Fr/I, where I is the moment of inertia. Why does it accelerate linearly, if the only force I'm applying is a torque? 
 A: 
Why does it accelerate linearly, if the only force I'm applying is a torque?

Forces and torques are separate things, but still related. A torque is produced by a force
$$\mathbf\tau=\mathbf r\times\mathbf F$$
So fundamentally you are just applying a force. That affects the linear motion of the object. If your force also happens to have a torque about the COM of the object, then the object will start spinning as well.
In other words, your force is not a torque. Your force has a torque about the COM. Torque is a manifestation of where and in what direction the force is being applied to the object.
Another way to look at this is to compare units. Forces have units of $\rm N$ where as torques have units of $\rm N\cdot\rm m$, so they cannot be the same thing.

Now, getting to how a tangential force causes a linear acceleration, it is easier to go more general. Let's say we have a rigid body of total mass $M$ consisting of $N$ particles. We know that the center of mass of the body is given by
$$r_{cm}=\sum_i^N\frac{m_i\mathbf r_i}{M}$$
Taking time derivatives, we can write the velocity and acceleration of the center of mass as well
$$v_{cm}=\frac{\text d}{\text dt}r_{cm}=\sum_i^N\frac{m_i\mathbf v_i}{M}$$
$$a_{cm}=\frac{\text d}{\text dt}v_{cm}=\sum_i^N\frac{m_i\mathbf a_i}{M}$$
Using Newton's second law we know that, for each particle
$$\mathbf F_i=m_i\mathbf a_i$$
and so putting it all togther we see that
$$a_{cm}=\sum_i^N\frac{\mathbf F_i}{M}$$
or rewriting things
$$\sum_i^N\mathbf F_i=Ma_{cm}$$
Now we need to think about the force we apply and the forces on each particle. The net force on any particle is the sum of any external forces acting on that particle and any forces between the other particles in the body:
$$\mathbf F_i=\sum_{ext}\mathbf F_i^{ext}+\sum_{j}^N\mathbf F_{j\rightarrow i}$$
Therefore:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}+\sum_i^N\sum_j^N\mathbf F_{j\rightarrow i}$$
Now, this double sum in the second term covers all $(i,j)$ pairs, so we can rewrite this term by double counting and then correcting for our double counting:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}+\frac12\sum_i^N\sum_j^N\mathbf (\mathbf F_{j\rightarrow i}+\mathbf F_{i\rightarrow j})$$
However, we know that by Newton's third law, $\mathbf F_{j\rightarrow i}=-\mathbf  F_{i\rightarrow j}$ So our "double count" sum is actually $0$. Therefore:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}=\mathbf F_{net}^{ext}$$
Therefore, bringing everything together:
$$\sum_i^N\mathbf F_i=\mathbf F_{net}^{ext}=Ma_{cm}$$
Notice how we have not assumed anything about the torques of these forces! So we see that, independent of what torques the applied forces might have, external forces alter the linear acceleration of the center of mass of the system of particles.
A: You're saying it: it's a rigid body. Inner distances are kept the same, so if one part suffers a force, the whole object suffers it. To be accurate, the force is invested in moving all particles, and the net effect is like if the whole force acted on the CoM.
A: Why does it accelerate linearly, if the only force I'm applying is a torque?
A torque $τ$ is simply the moment about a point created by the product of a force $F$ times the distance $r$ between the point and the force perpendicular to the moment arm.  So a torque cannot exist without one or more forces.  
To analyze the motion of the body subjected to forces with respect to a point in space you need to take the sum of the moments about the point, and the sum of the forces. If the sum of the moments and forces are zero, the body is in equilibrium.
If there is a net force applied to a rigid body, no mater where it is applied, it will experience a linear acceleration (translation). If that net force is not applied to the center of mass (COM) of the body, the body will experience, in addition to the linear acceleration, an angular acceleration (rotation) due to the moment (torque) about the COM. 
In order for the body to experience rotation without translation, you would need to apply what is called a “couple”. A couple is a system of two parallel forces that are equal in magnitude, and opposite in direction. The rotation occurs about a point midway between the parallel forces. Consequently you have a zero net force and there would rotation without translation.
The linear and angular accelerations described above will, of course, last only as long as the forces and torques are applied to the rigid body.
Hope this helps. 
A: My experience with SE is less than six months old. But I have already
learnt that the topic of rigid motion is the most frequently asked (perhaps only second to relativity), surely the less understood. It must be added that it's among the hardest and more complex in classical physics.
So no wonder if questions are often ill-posed and answers less
than satisfactory and always incomplete - to be complete one had to
write a book, and on the other hand one might ask himself what the
questioner would understand.
As an example your question can be analyzed. You write

Consider a rigid sphere in empty space [...] If a tangential force
  (torque) F (or any force not along the line of the centre of mass) is
  applied, it is my knowledge the sphere will translate and rotate. its
  COM will translate with an acceleration F/m, and it will rotate with
  an angular acceleration of Fr/I, where I is the moment of inertia. Why
  does it accelerate linearly, if the only force I'm applying is a
  torque?

About a force not being a torque, Aaron Stevens already answered. I
would only add that the english term "torque" is frequently misapplied
and misinterpreted. As I understand it, it's a term of common use in
mechanical engineering (e.g. the torque of a motor) but in my opinion
it would a good thing if it were not used in physics. I would prefer
the more lengthy expression "moment of a force" and in case more forces
are present "resultant moment".
Coming back to your question: what do you mean by "tangential force"?
Its meaning is not obvious because a force has an application point
and a direction. Tangential should mean that the application point is
on the sphere's surface and its direction is tangent to the surface. But
if the sphere rotates, what happens to the application point? Does it
remain fixed wrt the sphere? And what about the force's direction?
Does it remain the same or does it rotate together with the sphere? If your
question refers to a single instant of time, there is no problem. But
it's difficult not to consider what will happen at later times.
Anther problem arises about your assumption as to the ensuing motion.
Not for com's acceleration: Aaron has shown that it's always equal to
resultant of external forces divided total mass. (To @Aaron in some of
your formulas you missed boldface character for com position, velocity
and acceleration. Or was it intentional?)
This is rather simple, the simplest aspect of rigid motion. Much
harder to state and understand what a rotation will be if external
forces have a nonvanishing moment wrt com.
You speak of an angular acceleration $\mathbf{Fr/I}$ (boldface is
yours). First of all I would object to the boldface, generally used
for vectors whereas you use it or mass, radius, moment of inertia. But
your formula is the right angular acceleration? It is, for a
homogeneous sphere and for a strictly tangential force. The sphere is the simplest case as far as rigid motion in concerned, but you shouldn't think that what is true for a spherical body is also true for different shapes. You should also state the rotation axis: it's the straight line passing through com and perpendicular both to force and to the radius joining the sphere's center to the application point of force.
A final note about the case of a couple, examined by Bob D. He writes

The rotation occurs about a point midway between the parallel
  forces.

This isn't true. In case of a system of external forces of zero
total moment wrt com (a couple is a particular case) com motion is
unnaffected and rotation is always around an axis through com,
directed as the vector resultant moment.

The linear and angular accelerations described above will, of course,
  last only as long as the forces and torques are applied to the rigid
  body.

This is true as far as com motion is concerned. It's also true for a
spherical body: once there is no longer a resultant moment the sphere
will rotate uniformly. But a body of different form could experience,
even in absence of external forces, a much more complicated rotatory
motion.
