Current is the rate of flow of charges with time. When passing through each resistance in a series circuit, charges lose energy and "slow" down due to collisions with atoms (metal cations). When time increases, current decreases. I agree number of charges remains same but what about time?
Electric Circuits – Why Current Remains the Same in a Series Circuit
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Consider the following analogy to water pipes: Wires are like pipes already filled with water; the water resembles the movable charge, in case of metal electrons. Voltage is a pressure difference between two points, e.g. because one is higher up than the other. The pipes are really broad in comparison to the slow flow of water in it, so one can ignore friction/turbulence. Resistors are like chokes, simply a very narrow passage.
From this point of view it should be clear that the $\frac{volume\space of\space water}{time \space interval}$ is the same everywhere in the pipe. If you leave the pressure constant and enlarge the choke point, there will be more flow of water. If you squeeze it almost tight, there will be almost no flow of water, even though the pipes are nice and wide everywhere else.
From this point of view it should be really intuitive and trivial how simple circuits involving Rs and voltage sources behave.
My guess is that you confused current (which is $\frac{charge}{time \space interval}$ ) with the speed the charges move, which is called [drift velocity](see wikipedia).
This site has correct and very intuitive explanations of related concepts, and Ohm's Law and parallel circuits are very important but also easily understood given the analogy I gave above.
The amount of current in a series circuit is the same through any component in the circuit. This is because there is only one path for current flow in a series circuit.
This means that at any point along a series circuit, the current is the same. The current is still free to change, but if it changes, it changes everywhere. This just means that whatever the current may be, you will measure the same current at any point along a series circuit.
As for why adding resistance decreases current, you can apply Ohm's Law. For the same potential difference, you get $I=\dfrac VR$ so as $R$ increases, $I$ decreases.
Best Answer
Imagine a scenario. Take a battery having a potential of $V$. The ends of the battery is connected by a wire that is resistance less(an ideal approximation). In that case, the electrons in the wire are offered no 'resistance' and hence, they are free to move from the negative end to the positive end of the battery. From simple application of Ohm's law $$I=\frac{V}{R}$$ we see that the current is not defined, but in the limiting case of $R\rightarrow 0$, $I\rightarrow \infty$. Though the Ohm's law is an approximate law and we need a better treatment of the scenario to understand what is happening, we as of now know that the electrons don't travel through the resistance less wire instantaneously. There are electrons all along in the wire which just 'drift' towards the positive end of the battery without any obstruction and so quite fast(not instantaneously).
Now, if we include a resistor of resistance $R$ in this scenario, the drift of the electrons are no doubt restricted and this restriction of the motion of charges through the resister is indeed what causes the potential drop across the resistor, and the drift through the resistor occurs at a smaller rate, hence the smaller current.
For two resistors in series, this drift is slowed by both of the resistors(hence lesser current as compared to the single resistor case). However, the drift velocity through the entire wire has to remain the same as electrons can't accumulate anywhere in the wire if there are non regular speeds of their drifts as pointed out by @Nuclear Wang and the OP wrote in the comments. Also, electrons do lose energy due to collisions and that loss appears as heat that reduces the potential in due time and hence the current all in all deceases. But the current can't have various values in such a series circuit.