Basic thermodynamics question about work
Well, the answer is tricky. But before the really tricky part, you may want to take a step back and consider a general statement of the 1st law,

$$dU = \sum_i \left(\frac{\partial U}{\partial q_i} \right)_j dq_i$$

where $q_i$ is an extensive coordinate of the system. Note this is consistent with our standard description of the relation between generalized forces and energy (the derivatives wrt coordinates are forces).

In the case of pV work, $q_i = V$ and  

$$ \left(\frac{\partial U}{\partial V} \right)_j = -p $$ 

The partial derivative in question is evaluated holding all other extensive variables (including number of particles) constant. If there is a change in chemical composition of the system we can define a chemical force (although referred to as the chemical potential of substance i) as 

$$ \left(\frac{\partial U}{\partial n_i} \right)_j   = \mu_i  $$ 

and an infinitesimally small amount of chemical work as 

$$ \left(\frac{\partial U}{\partial n_i} \right)_j dn_i  = \mu_i dn_i $$ 

As noted in the comments, the formulation combining the first and second laws of thermodynamics presented in the OP is a particular representation of the first law for the case of reversible pV work, a more general formulation of the 1st law being 

$$\begin{align} dU &amp;= dq + dw\end{align}$$

Note that while $dU$ is exact, the differentials on the right-hand-side of the previous equation may be inexact. Distinguishing between pV and other types of work for a reversible process (for which the differentials $dw$ and $dq$ become exact) we can write

$$\begin{align} dU &amp;= dq_{rev} + dw_{pV,rev} + dw_{nonpV,rev}   \\ &amp;= dq_{rev} + dw_{pV,rev} + \sum_i \mu_i dn_i   \\ &amp;= TdS + dw_{pV,rev} + \sum_i \mu_i dn_i \end{align}$$

The second equality follows from the first using the 2nd law. Then since $dw=-p_{ext}dV$ in the case of pV work  and since $p_{ext} = p_{sys}=p$ when the work is carried out reversibly, you arrive at the expression in the OP. 


  I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.


You have to be careful how you define the process that alters the number of particles. Do you have an open system that allows the particles in from the surroundings? 

Assume you transfer an ideal gas at constant T from the surroundings into the system, which consists of a vessel at lower pressure with a small orifice which we seal at some point. Since the gas is ideal there is no change in energy with pressure at constant T. There is no mechanical work done, since there is no volume change, but the change in the number of particles clearly amounts to a transfer of energy. This is accounted for in the term $\sum_i \mu_i dn_i $ of $dU$, provided the entropy is constant.  Changing the number of particles amounts to "chemical" work.


  Secondly, in the same manner I also don't understand why dQ is defined as TdS and not as TdS+SdT. I would think that a change in temperature would surely change the internal energy.


A change in temperature can change the internal energy, but that does not require a change in the definition of heat or entropy. The equation $dQ=TdS$ holds only for a reversible transfer of heat, it is not general.

Now to the tricky part. 

You can write 

$$dU = -pdV-Vdp +SdT + TdS + \sum_i \mu_i dn_i + \sum_i n_i d\mu_i  $$

since 

$$U = -PV + TS + \sum_i n_i \mu_i$$

The thing is you can show that 

$$ 0 = -Vdp +SdT + \sum_i n_i d\mu_i  $$

This is known as the Gibbs-Duhem relation.