Why does everything go in the direction of positive entropy change?
This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?

Introduction

The first law of thermodynamics tells us that in any process energy is conserved, it can be transformed and converted into different forms, but the total amount is unchanged. 

The second law imposes limits on the efficiency of energy conversion processes. For a robust understanding of the second law it is necessary first to understand Carnot cycles. As such I will divide my answer into sections. Many of the points in the following are intricate in their application and a good knowledge of the first law is required. At the time classical thermodynamics was developed the main interest of the research was into the efficiency of heat engines. Historically these are quite interesting, but that is why you need to understand a Carnot cycle first which in essence is just a really efficient engine (the most efficient). 

Keep in mind that in any heat engine a supply of heat $Q$ can either be converted to work $W$ or internal energy $\Delta U$. Since an isothermal process means $\Delta U=0$ then the most efficient cycle will have two isotherms. However two isotherms do not constitute a cycle. The most efficient way to connect these two isotherms is with adiabatic processes (no heat exchange with the surroundings). With two adiabatic lines we finally have our cycle. Graphically to maximise the work you want to maximise the area under the cycle. 

By cycle we mean that a system returns to exactly the same state as before the change or process occurred. 

Carnot Cycles

Work can be obtained from an engine of there are heat sources at different temperatures, and heat will flow from a hot body to a cold body without work being performed until thermal equilibrium is attained. Carnot realised therefore that any temperature difference either produces work, or is wasted as spontaneous flows of heat. 

Lets use an ideal gas as our substance. The figure shows a Carnot cycle. It is reversible and has two reservoirs. Where $ab$ is an isotherm at temperature $T_1$ and heat $Q_1$ enters from a reservoir at $T_1$. The line $cd$ is an isotherm at a lower temperature $T_2$ where heat is given to another reservoir at $T_2$. The line $bc$ and $da$ are adiabatic processes. The work done $W$ is the area enclosed by the cycle. 

We might choose to represent the Carnot engine according to the figure below. Where $W$ is the work done by the engine. The efficiency $\eta$ is a measure of "what we get out for what we put in" so in this case the work output for the heat input. 
\begin{equation}
\eta =\frac {W}{Q_1}
\end{equation}
If we remember the first law and that $\Delta U=0$ then we may write $W=Q_1-Q_2$ denoting the sign convention for heat into the system and work done on the system to be positive and vice versa. 
\begin{equation}
\eta =1-\frac{Q_2}{Q_1}
\end{equation}
Kelvin-Planck statment

It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at a uniform temperature and the performance of an equivalent amount of work.

This basically says that some heat needs to be given to a body of colder temperature otherwise by the above equation we would have 100% efficiency. If you were to violate this you could just have perpetual motion machines or cars that just absorbed heat and produced work. It may be summarised as a process whose only effect is the complete conversion of heat into work is impossible. However the opposite process is allowed  where all the work is transferred to heat completely 

The Clausius statement

*It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. *

What this means is that work must be performed if heat is to be transferred form a colder to a hotter body. We might be more familiar with this statement if we called this engine a fridge. An engine that extracts heat from a cold body and delivers it to a hot body when work is performed. 

The kelvin-Planck statement is entirely equivalent to the Clausius statement. We will skip any proof or mathematics of why this is so. 

Carnot's Theorem

No engine operating between two reservoirs can be more efficient that a Carnot engine operating between those same two reservoirs.

The proof is as follows. Assume that such an engine did exist $E'$ with efficiency $\eta '$ that takes heat $Q_1'$ from the hot body and does work $W'$ delivering heat $Q_2'=(Q_1'-W')$ to the cold bath. 

Let us now imagine a Carnot engine $C$ with heats $Q_1$ and $Q_2$, efficiency $\eta$ adjusted in such a way that it performs the exact amount of work $W$ as the "more efficient engine $E'$ so that $W=W'$, and $Q_2=Q_1-W$. 

Iff,
\begin{equation} 
\frac{W'}{Q_1'}&gt;\frac{W}{Q_1}\ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ Q_1&gt;Q_1'
\end{equation}
The Carnot engine is reversible so it may be driven backwards, acting as a composite device with $E'$. It extracts positive heat $(Q_1-Q_1')$ form the cold reservoir and delivers the same heat to the hot reservoir with no external work being required. 

However since the reservoir is large the temperature is unchanged with the addition of this heat. This means that the Clausius statement is violated. Therefore $\eta'&gt;\eta $ is not true, and $E'$ can never exist. If $\eta '=\eta$ then $Q_1'=Q_1$ and no net heat transferred for no work. Therefore we say,
\begin{equation}
\eta' \leq \eta 
\end{equation} 
Entropy

Consider a substance that can do work under going a cycle so that at the end of its cycle its state is unchanged at $T_1$. Lets say we change the first state to infinitesimally close neighbouring state 2 at $T_2$ by adding a small amount of heat $\delta Q_1$ from . If we do this with a Carnot engine with the temperature baths of state 1 $T_1$ and some reservoir at $\bar T$ then $T_1$ supplies $\delta Q_q$ to the working substance while the Carnot engine supplies the same amount to the $T_1$ reservoir to leave it unchanged, in turn taking heat $\{\bar T/T_1\}\delta Q_1$ from a principle reservoir. 

So in short we have performed a change from states 1 to 2 with external work $\delta W_1$ and  extraction of heat $\{\bar T/T_1\}\delta Q_1$ from the principle reservoir. This is one system of a composite process of identical systems. The diagram really helps here! 

The composite system is assumed that $\Delta U=0$ and that the heat supplied to it is;
\begin{equation}
Q=\sum _i\frac{\bar T}{T_i}\delta Q_i
\end{equation}
Where the sum is over all the little systems making up the composite. The external work performed is;
\begin{equation}
W=\sum _i\delta W_i
\end{equation}
The first law is then $0=Q-W$. Looking at the second diagram below for the composite system we can see that this violates the Kelvin-Planck statement. The only way that this works is if work is done on the system and an equal quantity of work flows out. Or it works if both $W$ and $Q$ are zero. 
\begin{equation}
W=Q\leq 0
\end{equation} 
So 
\begin{equation}
\bar T\sum _i\frac{\delta Q_i}{T_i}\leq 0 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \sum _i \frac{\delta Q_i}{T_i}\leq 0
\end{equation}
If you take the limit of this then over one complete cycle we arrive at the Clausius inequality! 
\begin{equation}
\oint \frac{dQ}{T}\leq 0
\end{equation}
It is important to know that the $T$ appearing in the integral is the temperature of the heat supplying bath!

If the cycle is reversible then there is no reason why we couldn't do the proof in the other direction and hence obtain the inequality sign the other way around. This only holds for reversible cycles. The only way to satisfy both inequalities physically is if for reversible cycles we equal the integral to zero exactly. This means that for reversible cycles the integral is equal to zero!

The final step is to now use this Clausius inequality to prove the second law. Imagine a cycle that is completely reversible between two states $i$ and $f$ with two reversible paths $r_1$ and $r_2$ connecting them. 
\begin{equation}
\oint \frac{dQ_r}{T}=_{r_1}\int^{f}_{i} \frac{dQ_r}{T}+_{r_2}\int_{f}^{i}\frac{dQ_r}{T}=0
\end{equation}
Thus,
\begin{equation}
_{r_1}\int ^{f}_{i}\frac{dQ_r}{T}=_{r_2}\int^{f}_{i}\frac{dQ_r}{T}\end{equation}
What all this means is that the integral is path independent SO THERE MUST BE A PATH INDEPENDENT FUNCTION !! We call this state function entropy. 
\begin{equation}
S_f-S_i=_r\int ^f_i \frac{dQ_r}{T}
\end{equation}
For this equality to hold the heat exchanged must be defined over a reversible path! That is to say if the entire cycle is not reversible then we have to use the Clausius stamen again. Lets suppose that one off the paths is irreversible now.
\begin{equation} 
 \int^f_i\frac{dQ}{T_0}+_r\int ^i_f\frac{dQ_r}{T}\leq 0
\end{equation}
 With a bit of interval switching,
\begin{equation}
\int ^f_i\frac{DQ}{T_0}\leq -_r\int ^i_f\frac{dQ_r}{T}=_r\int ^f_{i}\frac{dQ_r}{T}=S_f-S_i
\end{equation}
And finally the way we normally write this is in an infinitesimal form,
\begin{equation}
\frac{dQ}{T_0}\leq dS
\end{equation}
This is more general than before since it includes the previous equation too, equality holds if the cycle is reversible. So , in an infinitesimal irreversible process between a pair of states the entropy change is $dS$, if the system is thermally isolated then $dQ=0$ and we generate the principle of increasing or "maximum" entropy. 
\begin{equation}
S_f-S_i \geq 0
\end{equation}
Or if you prefer $dS\geq 0$. The entropy of a thermally isolated system increases in any irreversible process and is unaltered in a reversible one.