Electronic – Transmission of serial data characters

To understand all of this stuff, take a look at "Table 18-1. Equations for Calculating Baud Rate Register Setting" from the datasheet on page 189. The equation you suggested you found in some example code

UBRR1 = (F_CPU / 4 / baud - 1) / 2;

... is kind of close to the equation for Asynchronous Double Speed mode (U2Xn = 1)... but not exactly.

If you know the baud rate and crystal speed you intend to operate at, I would just pull the register settings out of "Table 18-9. Examples of UBRRn Settings for Commonly Used Oscillator Frequencies" on datasheet page 210 and call it a day. The only reason to use any of those equations is if you want to be able to change the settings dynamically at run-time for some reason (or if you want to do thins "elegantly" in software, I prefer compile time certainty for something like this to remove doubt).

For the settings you outlined in your question, that would be:

UCSR1A = 0;                         // importantly U2X1 = 0
UCSR1B = 0;                         // interrupts enabled in here if you like
UCSR1C = _BV(UCSZ11) | _BV(UCSZ10); // no parity, 8 data bits, 1 stop bit
UCSR1D = 0;                         // no cts, no rts
UBRR1  = 103;                       // 9600 baud @ 16MHz XTAL with U2X1 = 0    

As a side note, while the hardware does store the baud rate register (UBBRn) as two 8-bit registers, in software there is no need to treat it as such. You can access the High and Low registers (e.g. UBRR1H and UBRR1L respectively) but you can also just assign to (or read from) the named "combined" register as though it was a 16-bit register.

UBRR1 = 0x0343; is functionally equivalent to UBRR1H = 0x03; UBRR1L = 0x43;

Simply put, you're not transmitting valid UART serial.

Serial comprises of one start bit, 7 or 8 data bits, 1 or 0 parity bits, and a stop bit.

In total that's 10 bits. The most common arrangement of that is 1 start, 8 data, and 1 stop.

Note that the UART has an "idle" state - HIGH in this case. The start bit is the opposite of the idle level, and the stop bit returns you to that level. The receiver waits for the transition from idle to not idle to know when it's starting to receive.

To transmit 16 bits of data you have to transmit it as two separate bytes. That would be 1 start, 8 data, 1 stop, 1 start, 8 data, and 1 stop. That's a total of 20 bits.

The baud rate is the number of symbols per second - that is the number of 1 or 0 bits, not the number of whole packets. So if you are shifting the data out at 195312.5Hz then the baud rate is 195312.5

Of course you now need a way of knowing which of your bytes are which - and now you are into the realms of data packets, and higher level protocols.